Equilibrium - Class 11 Medical Chemistry - Extra Questions

$$K_f=2.0$$

Add few drops of solution from all three test tubes on the red litmus paper separately. The solution which turns red litmus to blue contains basic solution. Use this blue litmus paper to test the solutions in other two test tubes. The solution from the test tube which turns blue litmus paper to red will be the acidic solution and solution of the test tube which do not change either red or blue litmus paper contain water.

$$\bullet$$ More acidic substance less is the pH.

$$\bullet$$ The large crystals of common salt found in underground deposits are called rock salt. Rock salt is mined from the underground deposits just like coal.

Compounds like Silver Chloride are light-sensitive and they may react to light very fast and lose their properties. That's why they are to be stored in dark bottles to prevent the entry of light.

$$[H_{3}O^{+}]$$= $$n \alpha C$$

Solubility is defined as the upper limit of solute that can be dissolved in a given amount of solvent at equilibrium. Factors on which solubility depends;

Ammonium hydroxide solution is weak electrolyte so it dissociates into ions very less and its solution contains both molecules and ions.

The ionisation constant of ammonia and ammonium hydroxide is same
$$\displaystyle NH_3 + H_2O \rightarrow NH_4^+ + OH^-  K_1  \frac {[NH_4^+][OH^-]}{NH_3}$$
$$\displaystyle NH_4OH \rightarrow NH_4^+ + OH^-   K_2 = \frac {[NH_4^+][OH^-]}{[NH_4OH]}$$
But $$\displaystyle [NH_3] = [NH_4OH] $$ (this is because when ammonia is dissolved in water, it is converted to ammonium hydroxide).
Hence, $$\displaystyle  K_1=K_2$$
Hence, the ionization constant of ammonia and ammonium hydroxide is same.

Acetic acid is less acidic in sodium acetate solution than in NaCl solution due to common ion effect of acetate ion.

Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.

Acids, in general, are sour to taste. 

$$\displaystyle pH = 9.95 $$, 

Strong acids dissociate completely, whereas weak acids dissociate only partially. The majority of the weak acid as remaining intact, which will be true at most normally used concentrations of acetic acid. If the same concentration of acetic acid and hydrochloric acid are taken, acetic acid furnishes less number of $$ H^+$$ is a strong acid in aqueous solution.

Hint: Acids are compounds that release $$H^+$$ ions easily.

$$ CH_3COOH   +   NaHCO_3     \rightarrow       CH_3COONa   +  CO_2   +   H_2O $$

Tooth enamel is made up of hard matter calcium phosphate. Food particles are degraded by bacteria in the mouth to form acid. The $$pH$$ drops below $$6$$. Due to this, phosphates corrode which causes tooth decay.

Calcium hypochlorite is known as bleaching powder its molecular formula is $$Ca(OCl)_2$$

Even clean, normal rain has a pH of about 5.6. This is because it reacts with carbon dioxide in the atmosphere and forms mildly acidic carbonic acid before it becomes rain. Therefore the statement is false.

When calcium hydroxide reacts with ammonium chloride, ammonia gas is liberated. Ammonia is a pungent smelling gas and is alkaline in nature.

Buffer solution : A buffer solution is one which maintains its pH fairly constant even upon the addition of small amounts of acid or base.
Two common types of bufffer solutions are : 
1. a weak acid together with a salt of the same acid with a strong base. These are called acid buffers, 
(eg): $$CH_3COOH+ CH_3COONa$$  
2. a weak base Mid its salt with a strong acid. These are called basic buffers,
(eg) : $$NH_4OH+NH_4Cl$$

For an acidic buffer solution made from  acetic acid and sodium acetate,  
$$\displaystyle pH = pKa + log \dfrac {[CH_3COONa]}{[CH_3COOH]}$$

For a solution containing equal concentrations of acetic acid and sodium acetate

$$\displaystyle \dfrac {[CH_3COONa]}{[CH_3COOH]} = 1$$
$$\displaystyle log \dfrac {[CH_3COONa]}{[CH_3COOH]} = log 1 =0$$
$$\displaystyle pH =pKa$$

Solubility of silver chloride decreases in the presence of sodium chloride because of common-ion effect.

Sodium chloride is a strong electrolyte and completely dissociates in aqueous solution to provide chloride ions that are common to silver chloride. This shifts the equilibrium for dissociation of silver chloride towards left.
$$\displaystyle AgCl \rightleftharpoons Ag^+ + Cl^-$$
$$\displaystyle NaCl \rightarrow Na^+ + Cl^-$$

$$NH_{4}Cl$$ is a strong electrolyte hence, dissociates completely, $$NH_{4}OH$$ is a weak electrolyte. Its dissociation is further suppressed by common ion $$NH_{4}^{+}$$ provided by $$NH_{4}Cl$$ in the solution.
$$NH_{4}Cl_{aq}\rightleftharpoons NH_{4_{aq}}^{+} + Cl_{aq}^{-}$$
$$NH_{4}OH_{aq} \rightleftharpoons NH_{4_{aq}}^{+} + OH_{aq}^{-}$$
This solution acts as a basic buffer and maintains its $$pH$$ around $$9.25$$. It is capable of resisting the change in $$pH$$ on addition of small amount of acid or alkali.
To Explain:
If a small amount of $$HCl$$ is added to this solution, $$H^{+}$$ ions of $$HCl$$ get neutralized by $$OH^{-}$$ ions already present and more of $$NH_{4}OH$$ molecules get ionized to compensate loss of $$OH^{-}$$ ions. Thus $$pH$$ practically, remains constant.
If a small amount of $$NaOH$$ is added to this solution, $$OH^{-}$$ ions of $$NaOH$$ combine with $$NH_{4}^{+}ions$$ already present in large number forming weakly ionized $$NH_{4}OH$$. Thus, $$pH$$ of solution remains practically unchanged.

pH stands for power of hydrogen, which is a measurement of the hydrogen ion concentration. The pH scale measures how acidic or basic a substance is. The total pH scale ranges from 1 to 14, with 7 considered to be neutral. A pH less than 7 is said to be acidic and pH greater than 7 is basic or alkaline.

Certain foods increase the acidity of the mouth, i.e. they drop the pH level. This includes sugary drinks, snacks, and some grains. The lower pH of the mouth creates an optimal environment for bad bacteria, which eventually causes tooth decay.

The formation of hydrogen ions and hydroxide ions form water is an endothermic process.
$${ H }_{ 2 }{ O }_{ (l) }\rightleftharpoons { H }_{ (aq) }^{ + }+{ OH }_{ (aq) }^{ - }\quad $$
The forward reaction absorbs heat. Thus according to Le Chatelier's principle increase in the temperature will shift the equilibrium to lower temperature. It means that the forward reaction will be favoured and more $${H}^{+}$$ and $${OH}^{-}$$ will be formed. Thus, the value of ionic product of water increases as temperature increases.
As $$pH=-\log[{H}^{+}]$$, thus with increase in temperature the pH of the neutral solution decreases.

Solutions which reserve acidity or alkalinity or which resists change of pH on the addition of small amounts of acid or alkali are called Buffer solution.
Prepared by addition of acid or base depending on which they are respectively called acidic and basic Buffer.
Basic Buffer : When a drop of strong acid say $$HCl$$ is added,
                           $$N{ H }_{ 4 }OH\rightleftharpoons N{ H }_{ 4 }^{ + }+O{ H }^{ - }$$
                           $$\begin{matrix} HCl & \rightleftharpoons  & { Cl }^{ - }+ & { H }^{ + } \\  &  & \upharpoonleft \downharpoonright  & \upharpoonleft \downharpoonright  \\  &  & N{ H }_{ 4 }Cl & { H }_{ 2 }O \end{matrix}$$
                                                Feebly ionised
The $$O{ H }^{ - }$$ ions produced from $$N{ H }_{ 4 }OH$$ combines with $${ H }^{ + }$$ ions produced from $$HCl$$ to form feebly ionized water molecule and therefore, the pH value does not change appreciably.

The three major buffer systems of our body are $$H_2CO_3/HCO_3^{\circleddash}$$ buffer system, phosphate buffer system and protein buffer system. Out of them, $$H_2CO_3/HCO_3^{\circleddash}$$ i.e. carbonic acid-bicarbonate buffer is the most important. Cellular respiration produces $$CO_2$$ and it is immediately converted to $$HCO_3^ {\circleddash}$$ in the blood plasma.

$$pH$$ of the soil indicates the acidity or alkalinity of the soil. Plants need a specific $$pH$$ range for their healthy growth. So to know that the $$pH$$ of the soil, samples of the same soil are collected from various places and are tested. This helps farmers to understand the acidic or basic nature of soil accordingly they can use substances to maintain a required $$pH$$ range.

Acidic substances have $$pH<7$$.

Let the solubility of $$Pb{I}_{2}$$ be $$S$$. Then
$$Pb{ I }_{ 2 }\rightleftharpoons { Pb }^{ 2+ }+2{ I }^{ - }$$
$$S$$             $$S$$           $$2S$$
Potassium iodide is a strong electrolyte and is completely ionised. It shall provide $${I}^{-}$$ ion concentration $$=0.1M$$
$$\left[ { Pb }^{ 2+ } \right] =S$$
$$\left[ { I }^{ - } \right] =(2S+0.1)M$$
$${ K }_{ sp }=\left[ { Pb }^{ 2+ } \right] { \left[ { I }^{ - } \right]  }^{ 2 }=S\times { \left( 2S+0.1 \right)  }^{ 2 }=4{ S }^{ 3 }+0.01S+0.4{ S }^{ 2 }$$
Neglecting $${S}^{3}$$ and $${S}^{2}$$
$$1.4\times { 10 }^{ -8 }=0.01S$$
or $$S=\cfrac { 1.4\times { 10 }^{ -8 } }{ 0.01 } =1.4\times { 10 }^{ -6 }mol\ { L }^{ -1 }$$

The equilibrium is:
(i) $$Ba{ SO }_{ 4 }\rightleftharpoons { Ba }^{ 2+ }+{ SO }_{ 4 }^{ 2- }$$
Let $$S$$ be the solubility in $$mol $$ $${Litre}^{-1}$$; then
$${ K }_{ s }=\left[ { Ba }^{ 2+ } \right] \left[ { SO }_{ 4 }^{ 2- } \right] ={ S }^{ 2 }$$
or $$1.5\times { 10 }^{ -9 }={ S }^{ 2 }$$
so, $$S=3.87\times { 10 }^{ -5 }mol\ { L }^{ -1 }$$
(ii) Let $$S'$$ be the solubility of $$Ba{ SO }_{ 4 }$$ in $$0.1M$$ $$Ba{Cl}_{2}$$ solution.
Total $${ Ba }^{ 2+ }$$ ions concentration $$=(S'+c)mol\ { L }^{ -1 }$$
and $${ SO }_{ 4 }^{ 2- }$$ ions concentration $$=S'mol\ { L }^{ -1 }$$
So, $${K}_{s}=(S'+c)S'=(S'+0.1)S'$$
or $$1.5\times { 10 }^{ -9 }=(S'+0.1)S'$$
or $${ S' }^{ 2 }+0.1S'=1.5\times { 10 }^{ -9 }$$
Neglecting $${ (S') }^{ 2 }$$
$$0.1S'=1.5\times { 10 }^{ -9 }$$
or $$S'=1.5\times { 10 }^{ -8 }mol\ { L }^{ -1 }$$

Let the degree of dissociation be '$$\alpha$$'
$${ CH }_{ 3 }COOH$$ is a weak electrolyte; thus
$${ \alpha  }^{ 2 }C={ K }_{ a }$$
$${ \alpha  }^{ 2 }\times 0.1=1.8\times { 10 }^{ -5 }$$
$$\alpha =1.34\times { 10 }^{ -2 }$$
$$\left[ { H }^{ + } \right] =\alpha C=1.34\times { 10 }^{ -3 }M\quad $$
$$pH=-\log { \left[ { H }^{ + } \right]  } =-\log { 1.34\times { 10 }^{ -3 } } =2.8729$$

The maximum concentration of $$\left[ { OH }^{ - } \right] $$ ions that will precipitate $$Mg{ \left( OH \right)  }_{ 2 }$$ is calculated by applying the equation
$${ K }_{ sp }=\left[ { Mg }^{ 2+ } \right] { \left[ { OH }^{ - } \right]  }^{ 2 }$$
$${ \left[ { OH }^{ - } \right]  }^{ 2 }=\cfrac { { K }_{ sp } }{ \left[ { Mg }^{ 2+ } \right]  } =\cfrac { 9.0\times { 10 }^{ -12 } }{ 0.05 } =1.8\quad \times { 10 }^{ -10 }$$
or $$\left[ { OH }^{ - } \right] =1.34\times { 10 }^{ -5 }M$$
$${ NH }_{ 3 }$$ is present in solution in the form of $${ NH }_{ 4 }OH$$
$${ NH }_{ 3 }+{ H }_{ 2 }O\rightleftharpoons { NH }_{ 4 }OH\rightleftharpoons { NH }_{ 4 }^{ + }+{ OH }^{ - }$$
$$0.05$$                        $$0.05$$
The ionisation of $${ NH }_{ 4 }OH$$ is suppressed by the addition of $${ NH }_{ 4 }Cl$$ (strong electrolyte)
$${ K }_{ { NH }_{ 3 } }={ K }_{ { NH }_{ 4 }OH }=\cfrac { \left[ { NH }_{ 4 }^{ + } \right] \left[ { OH }^{ - } \right]  }{ \left[ { NH }_{ 4 }OH \right]  } $$
Whole of the concentration of $${ NH }_{ 4 }^{ + }$$ ions is provided by $${ NH }_{ 4 }Cl$$.
$$\left[ { NH }_{ 4 }^{ + } \right] =\cfrac { { K }_{ { NH }_{ 4 }OH }\times \left[ { NH }_{ 4 }OH \right]  }{ \left[ { OH }^{ - } \right]  } =\cfrac { 1.8\times { 10 }^{ -5 }\times 0.05 }{ 1.34\times { 10 }^{ -5 } } =0.067M$$
i.e., $$\left[ { NH }_{ 4 }Cl \right] =0.067M\quad $$

$$pH=-\log { \left[ { H }^{ + } \right]  } $$
or $$\log { \left[ { H }^{ + } \right]  } =-pH=-5.2$$
$$\left[ { H }^{ + } \right] ={ 10 }^{ -5.2 }=6.3\times { 10 }^{ -6 }M$$
'$$\alpha$$' degree of dissociation$$=\cfrac { \left[ { H }^{ + } \right]  }{ C } =\cfrac { 6.3\times { 10 }^{ -6 } }{ 0.1 } =6.3\times { 10 }^{ -5 }$$
According to Ostwald's formula for weak electrolyte
$${ K }_{ a }={ \alpha  }^{ 2 }C=6.3\times { 10 }^{ -5 }\times 6.3\times { 10 }^{ -5 }\times 0.1=3.69\times { 10 }^{ -10 }$$

The strength of an acid is directly proportional to square root of dissociation constants of acids. So relative strength of the given acids are

$$NH_4OH\rightleftharpoons NH_4{+}OH^-$$

$$HA\rightleftharpoons H{+}A^-$$

Concentration of $${ N }_{ 2 }{ H }_{ 4 }$$
$$\left[ { N }_{ 2 }{ H }_{ 4 } \right] =\cfrac { 0.16 }{ 32 } \times \cfrac { 1000 }{ 500 } =0.01M$$
         $${ N }_{ 2 }{ H }_{ 4 }+{ H }_{ 2 }O\rightleftharpoons { N }_{ 2 }{ H }_{ 5 }^{ + }+{ OH }^{ - }$$
At equi. $$C(1-\alpha)$$              $$C\alpha$$   $$C\alpha$$
$${ K }_{ b }={ C\alpha  }^{ 2 }$$
$${ \alpha  }^{ 2 }=\cfrac { { K }_{ b } }{ C } =\cfrac { 4.0\times { 10 }^{ -6 } }{ 0.01 } =4\times { 10 }^{ -4 }$$
or $$\alpha =2\times { 10 }^{ -2 }$$i.e., $$2$$%

For weak electrolyte,

$${ H }_{ 2 }O\rightleftharpoons { H }^{ + }+{ OH }^{ - },\left[ { H }_{ 2 }O \right] =\cfrac { 1000 }{ 18 } =55.56mol\quad { L }^{ -1 }$$
$$\left[ { H }^{ + } \right] =\left[ { OH }^{ - } \right] =C\times \alpha $$
$${ K }_{ a }=\cfrac { \left[ { H }^{ + } \right] \left[ { OH }^{ - } \right]  }{ \left[ { H }_{ 2 }O \right]  } =C\alpha^2=55.56\times (1.8\times 10^{-9})^2=1.8\times 10^{-16}$$
$${ K }_{ w }=\left[ { H }^{ + } \right] \left[ { OH }^{ - } \right] =C\alpha\times C\alpha=(55.56)^2\times (1.8\times 10^{-9})^2=10^-14$$

$$CH_3COOH\rightleftharpoons H^{+}+CH_3COO^-$$

$$BOH\rightleftharpoons B^++OH^-$$

$$NH_4OH\rightleftharpoons NH_4{+}OH^-$$

(i) $$CaF_2= Ca^{2+}+2F^-$$

Degree of dissociation '$$\alpha$$' for weak acid=$$\sqrt { \cfrac { { K }_{ a } }{ C }  } $$

Degree of dissociation '$$\alpha$$' for weak acid=$$\sqrt { \cfrac { { K }_{ a } }{ C }  } $$

$$Ag_2CO_3 \rightleftharpoons \underset {2S}{2Ag^+}+\underset {S}{CO_3^{2-}}$$

(i) Hydrolysis of ester (ethyl acetate) begins slowly but becomes fast after sometime.
Initially catalytic amount of an acid is added to ethyl acetate. Hence, the hydrolysis reaction begins slowly. With passage of time, more and more acetic acid is formed during the hydrolysis. This acetic acid acts as a catalyst and increases the rate of the reaction.
$$\displaystyle CH_3COOC_2H_5 + H_2O \xrightarrow {H^+} CH_3COOH + C_2H_5-OH$$

(ii) The pH value of acetic acid increases on addition of a few drops of sodium acetate
Sodium acetate is a strong electrolyte and completely dissociates to provide acetate ions.
$$\displaystyle CH_3-COONa \rightarrow CH_3-COO^-  + Na^+$$
Acetic acid is weak electrolyte and partially dissociates.
$$\displaystyle CH_3-COOH \rightleftharpoons CH_3-COO^-  + H^+$$
The dissociation of sodium acetate provides acetate ions that are common to the dissociation of acetic acid. Due to common ion effect, the dissociation of acetic acid is suppressed. This decreases the hydrogen ion concentration and increases the pH of the solution.

Answer the following:
(1) The effect of temperature on ionic product of water ( $$\displaystyle K_w$$)
With increase in temperature, the extent of the dissociation of water increases. Due to this, the ionic product of water increases.
At $$\displaystyle 0 \: ^oC$$, $$\displaystyle K_w = 0.11 \times 10^{-14}$$
At $$\displaystyle 10 \: ^oC$$, $$\displaystyle K_w = 0.30 \times 10^{-14}$$
At $$\displaystyle 25 \: ^oC$$, $$\displaystyle K_w = 1.00 \times 10^{-14}$$
At $$\displaystyle 1000 \: ^oC$$, $$\displaystyle K_w = 7.50 \times 10^{-14}$$

(2) The value of the ionic product of water ($$\displaystyle K_w$$) remains unchanged if some acid is added to it.
When an acid is added to pure water, hydrogen ion concentration increases and hydroxide ion concentration decreases. But the product of the hydrogen ion concentration  and hydroxide ion concentration (i.e, ionic product of water) remains same.

$$MgSO_{ 4 }+2HCl\rightleftharpoons Mg{ Cl }_{ 2 }+{ H }_{ 2 }S{ O }_{ 4 }$$

$$3NaOH+Fe{ Cl }_{ 3 }\rightarrow 3NaCl+Fe\left( OH \right) _{ 3 }$$

White precipitate of barium sulphate forms.

(i) Addition of an excess of Ammonium hydroxide.

(i) At appropriate temperature (below 473 K) reaction is:

(i) $$CuC{ O }_{ 3 }+{ H }_{ 2 }S{ O }_{ 4 }\rightarrow CuSO_{ 4 }+{ H }_{ 2 }O+C{ O }_{ 2 }$$

$$\underset{\text{At equillibrium}}{\underset{\text{Initially}}{}} \underset{\left( 1 - x \right)}{\underset{1}{C{H}_{3}COOH}} \longrightarrow \underset{x}{\underset{0}{{C{H}_{3}COO}^{-}}} + \underset{x}{\underset{0}{{H}^{+}}}$$

Reaction of $$CaCl_2$$ and $$Na_2CO_3$$

Common ion effect is a phenomenon in which degree of dissociation of any weak electrolyte is supressed by addition of small amount of strong electrolyte containing a common ion.

More is the acidic substance less is the $$pH$$

$$\bullet$$ Silver nitrate is a salt which decomposes and forms silver metal.

When milk changes into curd (yogurt), its pH value decreases. This is because during curd formation, lactic acid is produced which makes it acidic.

$$ Ca(OCl) Cl +H_2O \longrightarrow Ca(OH)_2 +Cl_ 2 $$

Moles of $$\mathrm{NaHSO_4}$$ present in 1.8 gm of $$\mathrm{NaHSO_4}$$ = $$\mathrm{\cfrac{1.8}{23+1+32+64} }$$ = $$0.015$$ moles = 15 milli moles

Hence the answer is 1.

If the soil is too acidic, then it can be treated with materials like quicklime or slaked lime as these materials are basic in nature and hence react with the excess acids present in the soil to reduce its acidity.

When calcium hydroxide is treated with chlorine gas it yields bleaching powder. It is shown in the following chemical reaction.

The electrolysis of an aqueous solution of sodium chloride gives us three products. They are sodium hydroxide, chlorine gas and hydrogen gas.

Three chemicals made from common salt are Sodium hydroxide, sodium carbonate and sodium hydrogen carbonate.

When chlorine and sodium hydroxide being produced during the electrolysis of brine are allowed to mix, a new chemical is formed is Sodium hypochlorite, $$(NaClO)$$. It is used in making household bleaches and for bleaching fabrics.

$$Calcium$$ $$Sulphate$$  is a salt whose solubility decreases with temperature.

(i) Plants require a specific pH range for their healthy growth. By using a large number of pesticides and fertilizers, pH of the soil changes, which makes it more acidic or basic. So, in the long run, the soil becomes infertile. This leads to soil erosion, damaging the environment also. So, use of these pesticides and fertilizers should be restricted.
(ii) The chemical pesticides sprayed by the farmers in their fields, stick to the fruits and vegetables. So, they should be removed by washing properly, before eating.
(iii) Care towards society, unselfishness

$$Potassium$$ $$bromide$$ and $$potassium$$ chloride$$.

(i) We should brush our teeth twice a day as bacteria present in the mouth, produce acids by degradation of sugar and food particles remaining in the mouth after eating.
(ii) Toothpaste is better than water for cleaning teeth because toothpaste is basic in nature. So this can neutralize the excess acid and prevent tooth decay.

On exposure to air, table salt $$(NaCl)$$ turns moist and ultimately forms a solution especially during rainy season because it contains impurities like magnesium chloride and calcium chloride which are deliquescent. Sodium chloride is not deliquescent.

Hint: When an ant bites, it injects the acidic liquid into the skin, which causes a burning sensation. It is neutralized by rubbing any basic salt on it.

If the farmer mixes a small amount of slaked lime solution or quick lime solution to the soil, acidic soil will convert into neutral soil. 

Common salt is odd because on heating, there is no change in color of compound. But in rest of the compounds, there is change is colour. 

It is product of concentration of $$[H_3O^+]$$ and $$[OH^-]$$ at a specific temperature.

The equilibrium between ions and unionized molecules is called ionic equilibrium.

The solutions which resist change in pH on dilution or with the addition of alkali or acid in small amounts are called buffer solution.

Lemon, oranges and tamarind taste sour because they have acids present in them and we know that acids have a sour taste. Lemons and oranges contain citric acid, while tamarind contains tartaric acid.

A. NaCl is basic salt (2).

Take a clean test tube and add solid $$NaCl$$ in it. 

Salt is an ionic compound formed in a neutralization reaction that has a high solubility in water.

Salts are electrically neutral. The sum of the charge of the positive ions and negative ions in a salt will be zero.

Plants require a specific $$pH$$ range for their healthy growth. To find the $$pH$$ value of the soil related to different crops, we can conduct an activity.

Lime juice - Citric acid
Curd - Lactic acid
Tamarind - Tartaric acid
Vinegar - Acetic acid.

$$H$$ atom is responsible for the common properties of acids.

acidic buffer
basic buffer
neutral buffer.

The $$pH$$ of soil is an important factor for crops. It is important to identify whether the soil of a region is suitable for a particular crop. Acidic soil is suitable for some crops while basic soil for a few others.

$$KCl \rightarrow K^{+} + Cl^{-}$$
$$HNO_{3} \rightarrow H^{+} + NO_{3}^{-}$$
$$Mg(OH)_{3} \rightarrow .Mg^{2+} + 2OH$$
$$H_{2}SO_{4} \rightarrow 2H^{+} + SO_{4}^{2-}$$
$$NH_{4}Cl \rightarrow NH_{4}^{+} + Cl^{-}$$
$$CuSO_{4} \rightarrow Cu^{2+} + SO_{4}^{2-}$$.

Consider basic buffer solution,
$$NH_4OH \overset{\rightharpoonup}{\leftharpoondown} \ NH_4^+ + OH^- ; NH_4Cl \longrightarrow \ NH_4 + Cl^-$$
Here, ammonium hydroxide is a weak electrolyte, in it's solution there exist equilibrium between it's ions and molecules.
Where ammonium chloride completely dissociates into it's ions.
Therefore, buffer solution contains large number of $$NH_4^+$$ ions followed by $$Cl^- , OH^-$$ and $$NH_4OH$$.

Buffer solution : The solution which resists change in pH on dilution or with the addition of small amounts of acid or alkali is called buffer solution. 
Properties of buffer solution 

• Buffer solution resists change in pH.
• It is used to prepare a buffer solution of particular pH. 

pH of Acidic buffer 
Consider a reaction,
$$HA \rightleftharpoons H^+ + A^-$$
$$K_a = \frac{[H^+]}{[A^-]}{[HA]}$$                     ......(i)
Where $$K_a$$ = Dissociation constant of acid 
Taking logarithm of equation ...(ii) 
This equation can be rewritten as :
$$-pK_a  = -pH + log \frac{[A^-]}{[HA]}$$
Reamanging the negative sign gives the final version of the Henderson - Hasselbalch equation 
$$pH  = pK_a + log \frac{[A^-]}{[HA]}$$             ........... (iii)
Examples of Simple buffere 

• $$CH_3COONH_4$$
• $$(NH_4)_2CO_3$$

(i) Since vinegar (acetic acid) is used to heal or neutralize the effect of wasp stings this means that the chemical present in the stings must be some base.
(ii) If there is no baking soda in the house then washing soda can be used to treat bee sting.

$$K_p=p_{H_2O}=23\ mm$$

A compound whose aqueous solution or melt conducts electricity called an electrolyte.

It converts any $$NH_{4}HCO_{3}$$ present into $$(NH_4)_2CO_3 $$.

The word acid comes from the Latin word acere which means sour.

Ionic product of water depends only on temperature. So, it remains unchanged when some acid is added into water.

The following are the ionic equations to show the presence of ions in aqueous solution:

$$\underset { 1\\ 1-\alpha  }{ HA } \rightleftharpoons \underset { 0\\ \alpha  }{ { H }^{ + } } +\underset { 0\\ \alpha  }{ { A }^{ - } } \quad $$

Now,

$$$K=c \times (\alpha) ^2 = c' \times (\alpha ') ^2$$  .... (1)

$$[{H}^{+}]=\left[ \cfrac { { K }_{ w } }{ \left[ { H }^{ + } \right]  } +\cfrac { { K }_{ a }\cdot Co }{ \left[ { K }_{ a }+{ H }^{ + } \right]  }  \right]$$
Substitute values in the above expression.
$$10^{-3}=\dfrac {10^{-14}}{10^{-3}}+ \dfrac {K_a \times 10^{-1}}{K_a+10^{-3}}$$
$$\dfrac {K_a}{K_a+10^{-3}}=10^{-4}$$
$$1+ \dfrac {10^{-3}}{K_a} = 10^{4}$$
$$K_a=10^{-7}=10^{-X}$$
$$\implies X=7$$

(A) At the start of the titration, the solution is basic due to hydrolysis of carbonate ion $$\displaystyle CO_3^{2-}  $$.

The molar mass of hydrazine is $$32$$ g/mol. $$0.16$$ g of hydrazine in $$500$$ mL corresponds to molarity of $$\dfrac {0.16}{32 \times 0.5}=0.01M$$

	$$N_2H_4$$	+ $$H_2O \to$$        $$N_2H_5^+$$	$$OH^-$$
Initial molarity (M)	$$0.01$$	$$0$$	$$0$$
Equilibrium (M)	$$001-x$$	$$x$$	$$x$$

The expression for base dissociation constant $$K_b$$ is $$K_b=\dfrac {[N_2H_5^+][OH^-]}{[N_2H_4]}$$

A buffer solution has limited capacity to absorb $${H}^{+}$$ or $${OH}^{-}$$ ion.

For weak acid HA : $$\alpha_{HA} = \displaystyle \frac{1}{100}=0.01, [HA]=0.1 \, M$$

Given  $${ K }_{ w }={ 10 }^{ -14 }$$

When we add titrate sodium carbonate solution (in beaker) with hydrochloric acid, we observe the following:
(A) At the start of titration, hydrolysis of $${CO_3^{2-}}$$ occurs.
(B) Before the first equivalent point, buffer solution of $${HCO_3^{-}}$$ and $${CO_3^{2-}}$$ is obtained.
(C) At the first equivalent point, amphiprotic anion is obtained with  $$pH=1/2 {(pK_{a1}+pK_{a2})}$$.
(D) Between the first and second equivalent points, buffer solution  of $${H_2CO_3}$$ and $${HCO_3^{-}}$$ is obtained.

The acid dissociation equilibrium is as given below:

The balanced chemical equation for the ionization of phenol is: 
$$\displaystyle PhOH + H_2O \rightleftharpoons PhO^- + H_3O^+$$.

For the titration between strong acid and weak base equivalence point occurs at ph 6 which is acidic in nature and methyl red is red in pH under 4.4 and yellow in pH over 6.2.

$$\displaystyle [H^+] = c \alpha = 0.1 \times 0.132 = 0.0132 M $$

Let c M be the initial concentration of $$\displaystyle [C_6H_5NH_3^+] $$ and x be the degree of ionization.
$$\displaystyle \underset {c(1-x)}{PhNH_2} + H_2O \rightleftharpoons \underset {cx}{PhNH_3^+} + \underset {cx}{OH^-} $$
$$\displaystyle K_b = \frac {[PhNH_3^+][OH^-]}{[PhNH_2]}=\frac {(cx)(cx)}{c(1-x)} = \frac {cx^2}{1-x} $$
But x is very small. Hence, $$\displaystyle 1-x \approx 1 $$.
Hence, $$\displaystyle K_b = cx^2 $$.
The degree of ionization of aniline, $$\displaystyle x = \sqrt {\frac {K_b}{c}}  = \sqrt {\frac {4.3 \times 10^{-10}}{0.001}}=6.56 \times 10^{-4}$$

$$\displaystyle [OH^-] = cx = 0.001 \times 6.56 \times 10^{-4} = 6.56 \times 10^{-7}M $$
$$\displaystyle [H^+] = \frac {K_w}{[OH^-]} = \frac {10^{-14}}{6.56 \times 10^{-7}} = 1.52 \times 10^{-8} $$
$$\displaystyle pH = -log [H^+] = -log 1.52 \times 10^{-8} =7.818 $$

$$\displaystyle  K_a$$ (conjugate base of aniline) $$\displaystyle = \frac {K_w}{K_b} = \frac {10^{-14}}{4.3 \times 10^4} = 2.32 \times 10^{-5} $$

$$\displaystyle pH = - log [H^+] = 4.15 $$

$$\displaystyle pK_a = -log K_a  = 4.74 $$
$$\displaystyle K_a = 10^{-pKa} = 10^{-4.74}= 1.8 \times 10^{-5} $$
Let x be the degree of dissociation. The concentration of acetic acid solution, C = 0.05 M
The degree of dissociation, $$\displaystyle x = \sqrt {\frac {K_a}{C}} = \sqrt {\frac {1.8 \times 10^{-5}}{0.05}} = 0.019 $$


(a) The solution is also 0.01 M in HCl.
Let x M be the hydrogen ion concentration from ionization of acetic acid. The hydrogen ion concentration from ionization of HCl is 0.01 M. The total hydrogen ion concentration
$$\displaystyle [H^+]  = 0.01+x$$
The acetate ion concentration is equal to the  hydrogen ion concentration from ionization of acetic acid. This is also equal to the concentration of acetic acid that has dissociated.
$$\displaystyle [CH_3COO^-] = x$$
$$\displaystyle [CH_3COOH] = 0.05-x $$
$$\displaystyle K_a = \frac {[H^+][CH_3COO^-]}{[CH_3COOH]} $$
$$\displaystyle 1.8 \times 10^{-5} = \frac {(0.01+x)x}{(0.05-x)} $$.....(1)
As x is very small,  $$\displaystyle 0.01+x \approx 0.01 $$
$$\displaystyle 0.05-x \approx 0.05 $$
Hence, the equation (i) becomes
$$\displaystyle 1.8 \times 10^{-5} = \frac {0.01x}{0.05} $$
$$\displaystyle x = 9.0 \times 10^{-5} M $$
The degree of ionization is $$\displaystyle \dfrac {[CH_3COO^-]}{[CH_3COOH]}= \frac {x}{c} = \frac {9.0 \times 10^{-5}}{0.05} = 1.8 \times 10^{-3} = 0.0018$$.

(b) The solution is also 0.1 M in HCl.
Let x M be the hydrogen ion concentration from ionization of acetic acid. The hydrogen ion concentration from ionization of HCl is 0.01 M. The total hydrogen ion concentration
$$\displaystyle [H^+]  = 0.1+x$$
The acetate ion concentration is equal to the  hydrogen ion concentration from ionization of acetic acid. This is also equal to the concentration of acetic acid that has dissociated.
$$\displaystyle [CH_3COO^-] = x$$
$$\displaystyle [CH_3COOH] = 0.05-x $$
$$\displaystyle K_a = \frac {[H^+][CH_3COO^-]}{[CH_3COOH]} $$
$$\displaystyle 1.8 \times 10^{-5} = \frac {(0.1+x)x}{(0.05-x)} $$.....(1)
As x is very small,  $$\displaystyle 0.1+x \approx 0.1 $$
$$\displaystyle 0.05-x \approx 0.05 $$
Hence, the equation (i) becomes
$$\displaystyle 1.8 \times 10^{-5} = \frac {0.1x}{0.05} $$
$$\displaystyle x = 9.0 \times 10^{-6} M $$
The degree of ionization is $$\displaystyle  \dfrac {[CH_3COO^-]}{[CH_3COOH]}= \frac {x}{c} = \frac {9.0 \times 10^{-6}}{0.05} = 1.8 \times 10^{-4} = 0.00018$$.

The degree of ionization, $$\displaystyle y = \sqrt {\frac {K_b}{c}} = \sqrt {\frac {5.4 \times 10^{-4}}{2 \times 10^{-2}}}=0.164$$
Let x be the amount of dimethyl aniline dissociated in presence of 0.1 M NaOH.
$$\displaystyle (CH_3)_2NH + H_2O \rightleftharpoons (CH_3)_2NH_2^+ + OH^- $$
The equilibrium concentrations are:
$$\displaystyle [(CH_3)_2NH] = 0.02-x \approx 0.02 $$
$$\displaystyle [(CH_3)_2NH_2^+] = x $$
$$\displaystyle [OH^-] = 0.1+x \approx 0.1 $$
The equilibrium constant expression is: 
$$\displaystyle K_b = \frac {[(CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]} = \frac {x \times 0.1}{0.02}$$
$$\displaystyle x = 1.08 \times 10^{-4} $$
The percentage of dimethylamine ionized is $$\displaystyle \frac {1.08 \times 10^{-4}}{0.02} \times 100 = 0.54 $$%.

After mixing, the concentration of $$\displaystyle NaIO_3 $$ is $$\displaystyle \frac {0.002}{2} = 0.001 M $$.
After mixing, the concentration of $$\displaystyle Cu(ClO_3)_2 $$ is $$\displaystyle \frac {0.002}{2} = 0.001 M $$.
$$\displaystyle NaIO_3 \rightleftharpoons Na^+ + IO_3^- $$
$$\displaystyle [IO_3^-] = 0.001 M $$
$$\displaystyle Cu(ClO_3)_2 \rightleftharpoons Cu^{2+} + 2ClO_3^- $$
$$\displaystyle [Cu^{2+}] = 0.001M $$
The ionic product of $$\displaystyle Cu(IO_3)_2 $$ is
$$\displaystyle [Cu^{2+}] [I^-]^2 = 0.001 \times 0.001^2 = 1 \times 10^{-9}$$
As the ionic product is less than the solubility product, no precepitation will occur.

The reduction of the degree of dissociation of a salt by the addition of a common-ion is called the common ion effect.
E.g.: In a saturated solution of silver chloride, we have the equilibrium:
$$AgCl_{(aq)}\rightleftharpoons Ag^{+}+Cl^{-}$$
When sodium chloride is added to the solution the concentration of $$Cl^-$$ ions will increases. The equilibrium shown above will be shifted to the left to form more of solid $$AgCl$$. Thus, the solubility of $$AgCl$$ will decrease.
 

(a) A buffer solution is one which maintains its pH fairly constant even upon the addition of small amounts of acid or base. 
e.g., for acidic buffer $$CH_3COOH + CH_3 COONa$$ 
basic buffer $$NH_4 OH + NH_4Cl$$.
Buffer action: Let us consider the common buffer system containing acetic acid and sodium actetate.
$$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$$
$$CH_3 COONa \rightarrow CH_3COO^- + Na^+$$
Since, the salt is completely ionised, it provides the common ions $$CH_3COO^-$$ in excess. The common ion effect suppresses the ionisation of acetic acid. This reduces the $$[H^+]$$ and pH of the solution is raised.
The pH of a buffer solution containing equimolar amounts (0.10 m) of acetic acid and sodium actetate is 4.74.
Addition of HCI: Upon the addition of HCI, the decrease of $$H^+$$ ions is counteracted by association with the excess of acetate ions to form unionised $$CH_3COOH$$. Thus, the added $$H^+$$ ions are neutralised and the pH of the buffer solution remains unchanged.
Addition of NaOH: When NaOH is added to the buffer solution, the additional OH ions combine with $$CH_3COOH$$ to give $$CH_3COO^-$$ and $$H_2O$$. Thus, pH of the buffer solution is maintained almost constant.

$$PbS \rightleftharpoons Pb^{2+}+S^{2-}$$         $$ZnS \rightleftharpoons Zn^{2+}+S^{2-}$$

Let solubility of $$AgCNS$$ and $$AgBr$$ in $$a$$ and $$b$$ respectively.
$$AgCNS\rightleftharpoons { Ag }^{ + }+CN{ S }^{ - }$$
                      $$a$$         $$a$$
 

$$AgCl \rightleftharpoons Ag^ + Cl^-$$

$$AgCl\rightleftharpoons { Ag }^{ + }+{ Cl }^{ - }$$
On adding ammonia solution, complex formation takes place
$${ Ag }^{ + }+2{ NH }_{ 3 }\rightleftharpoons \left[ Ag{ \left( { NH }_{ 3 } \right)  }_{ 2 }^{  } \right] ^{ + }...(i)$$
Where, $$x=$$ solubility of $$AgCl$$ in $${NH}_{3}$$
$$y=$$ amount of complex formed
$${K}_{sp}$$ of $$AgCl=\left[ { Ag }^{ + } \right] \left[ { Cl }^{ - } \right] $$
$$\quad 1.8\times { 10 }^{ -10 }=(x-y)\times x...(ii)\quad $$
$${K}_{c}$$ for equation (i) $$=\cfrac { \left[ Ag{ \left( { NH }_{ 3 } \right)  }_{ 2 }^{  } \right] ^{ + } }{ \left[ { Ag }^{ + } \right] { \left[ { NH }_{ 3 } \right]  }^{ 2 } } $$
$$\cfrac { 1 }{ 6.2\times { 10 }^{ -8 } } =\cfrac { y }{ (x-y)1 } ....(iii)$$
On solving euations (ii) and (iii) we get
$$y=0.0539M$$

During salt analysis, $$NH_4Cl$$ is added in sufficient quantity before adding $$NH_4OH$$ because otherwise the cations of higher group may get precipitated in group III.

$$NH_4Cl$$ is strong electrolyte, it decomposes completely. While the $$NH_4OH$$ is weak base it does not ionises completely. Thus due to presence of common ion $$NH^{4+}$$ in $$NH_4Cl$$, it supresses the ionisation of weak base $$NH_4OH$$ in order to decrease the $$OH^{-}$$ concentration so that higher group cations will not get precipitated.This process is called common ion effect.

$$\displaystyle [H^+]=0.001158 \ M$$
$$\displaystyle pH=2.94$$

The equilibrium constant $$\displaystyle K_P= \dfrac {P_A \times P_B}{P_{AB}}$$

Ethylenediamine $$(en)$$ is dibasic base, the 1st ionisation is quite feasible. Consideration of 1st ionisation is sufficient in order of calculate the $$pH$$ as $${ K }_{ 2 }$$ is $$1000$$ smaller than $${ K }_{ 1 }$$ and we can treat en as monobasic base.

1.The variation of molar conductance with dilution can be explained on the basis of degree of dissociation with dilution and not over concentration.

2.As the number of ions furnished by an electrolyte in solution depends upon the degree of dissociation with dilution.

3. With the increase in dilution, the degree of dissociation increases and as a result molar conductance increases. 

Antacids neutralize or reduce excess stomach acid to relieve burn, sour stomach, acid indigestion, and stomach upset they can also be used to relieve the pain of the stomach and duodenal.

Hence the correct answer is 4.

$$ CaOCl_2 +2CH_3COOH +2KI \longrightarrow ( CH_3COO)_2Ca+2KCl +I_2 +H_2O $$

$$H_2O+H_2O\leftrightharpoons H_3O^+\,+ OH^-$$

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2.5 mol of $$ KH_2PO_ 4$$ and $$ K_2HPO_4 $$ are present in a solution. So using  pH = $$pKa_2 + log [ \frac {salt}{acid} ] $$
But [Salt] = [acid hence pH = $$ pKa_2 $$
at 25 ml of KOH -
2.5 m mol of $$ KH_2PO_4$$ and 2.5 m mol of $$ H_3PO_4 $$ are present in a solution. so using ph - $$ pKa_1 + log  [ \frac {salt}{acid} ] $$
But [ Salt] = [acid ] hence pH = $$ pKa_1 $$
at 150 ml of KOH- 
$$ K_3PO_4 $$ salt are present in a solution  due to which salt hydrolysis will take place hence 
pH = $$  7 + \frac {1}{2} [ pK a_3 + log C ] $$
At 100 ml of KOH -
$$ K_2HPO_4 $$ are present in a solution which work as amphoteric in a solution so, pH = $$ \frac {pK a_2 +pK a_3 }{2} $$

$$\text{Bleaching powder does not dissolve completely in water because bleaching powder is actually a quite complex}$$ $$\text{ molecule.  It contains various other heavy calcium salts in it like calcium chloride, calcium hydroxide, and calcium}$$ $$\text{hypochlorite. These salts are insoluble in water. Hence, bleaching powder does not dissolve completely in water. }$$

$$\begin{array}{|l|l|l|l|}\hline  {\text { Name of the salt }} & & {\text { Salt obtained from }} \\\& \text { Formula } & \text { Base } & \text { Acid } \\\hline \begin{array}{l}\text { Ammonium } \\\text { chloride }\end{array}  & \mathrm{NH}_{4} \mathrm{Cl} & \mathrm{NH}_{4} \mathrm{OH} & \mathrm{HCl} \\\hline \text { Coppersulphate } & \mathrm{CuSO}_{4} & \mathrm{Cu}(\mathrm{OH})_{2} & \mathrm{H}_{2} \mathrm{SO}_{4} \\\hline \text { Sodium Chloride } & \mathrm{NaCl} & \mathrm{NaOH} & \mathrm{HCl} \\\hline \text { Magnesium nitrate } & \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2} & \mathrm{Mg}(\mathrm{OH})_{2} & \mathrm{HNO}_{3} \\\hline \begin{array}{l}\text { Potassium } \\\text { sulphate }\end{array} & \mathrm{K}_{2} \mathrm{SO}_{4} & \mathrm{KOH} & \mathrm{H}_{2} \mathrm{SO}_{4} \\\hline \text { Calcium nitrate } & \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} & \mathrm{Ca}(\mathrm{OH})_{2} & \mathrm{HNO}_{3} \\\hline\end{array}$$

$$Dimethylamine$$	$$>Ammonia$$	$$>Pyridine$$	$$>Urea$$
$$K_{b}$$       $$5.4\times10^{-4}$$	$$1.77\times10^{-5}$$	$$1.77\times10^{-9}$$	$$1.3\times10^{-14}$$

Higher the value of $$K_{b}$$ stronger will be the base. Among the given bases, the strongest base is Dimethylamine.

As we know that alkalis react with oil to form soap. As our skin contains oil so when we touch strong alkalis, a reaction takes place and the soapy solution is formed. Hence we should wear gloves.

Barium sulphate($$BaSO_4$$) and lead sulphate($$PbSO_4$$)

Silver chloride $$(AgCl_2)$$

$$Lead$$ $$chloride$$  is insoluble in cold water but dissolves in hot water.

Autoprotolysis of $$H_{2}O $$ takes place as follows:
   $$H_{2}O + H_{2}O \rightleftharpoons  H_{3}O^{+} + OH^{-} $$ 

Ionic product of water $$(K_{w})$$ increases with increases of temperature because $$K_{w} = [H_{3}O^{+}] [OH^{-}] $$ and dissociation of $$H_{2}O $$ to give $$H_{3}O^{+}$$ and $$OH^{-}$$ increases with increase of temperature.

The solubility product of $$Mg(OH)_2 $$ is high. Presence of $$NH_{4}Cl$$ suppresses the dissociation of $$NH_{4}OH $$ due to common ion effect thus giving low concentration of $$[OH^{-}]$$. The ionic product, therefore cannot exceed the solubility product.

This is done to convert $$NH_{4}HCO_{3}$$ usually present in large amounts along with $$(NH_{4})_2CO_{3} $$ to $$(NH_{4})_2CO_{3}$$ 
   

The strength of an acid depends upon the extent of ionization. Acetic acid is highly soluble in water but it dissociates partially in the aqueous solution to produce a small amount of $$ H^+ $$  ions and therefore. considered as a weak acid. 

$$pH$$ is a scale used to specify the acidity or basicity of an aqueous solution. The range lies within $$0 - 14$$, with being neutral at $$pH$$ = $$7$$.