Both proton donor and proton acceptor

Neither proton donor, nor proton acceptor

MCQ Questions for Class 11 Medical Chemistry Equilibrium Quiz 2

Acids when dissolved in ___________ dissociates into ions.

0%
juice
0%
water
0%
kerosene
0%
liquid

Explanation

When acids are dissolved in water, they dissociate into an anion and hydrogen ion. These hydrogen ions combine with water molecules to form hydronium ions and these ions are responsible for the acidity of the acids.

$H^++H_2O \rightarrow H_3O^+$

Hence, the correct answer is option (B).

The formula of hydroxyl ion is:

0%
$OH^+$
0%
$OH^-$
0%
$H^+$
0%
none of the above

Explanation

Some of the general properties of bases are as follows:

They are bitter to taste, turns red litmus paper blue and releases hydroxide ion in the solution medium.

The hydroxide ion that is released is responsible for the basic character of bases.

The formula for hydroxyl ion is

$(OH^-)$ .

Hence, the correct answer is option (B).

Acids are bitter in taste.

0%
True
0%
False

X is a colourless gas with a distinct odour which consist of nitrogen and hydrogen atoms. It is produced naturally in the human body and in nature. Which gas is it? Is it acidic or basic in nature?

0%
Ammonia, Acidic
0%
Hydrogen Chloride, Acidic
0%
Ammonia, Basic
0%
Hydrogen Chloride, Basic

Nitric acid is acidic in nature.

0%
True
0%
False

Curd contains tartaric acid.

0%
True
0%
False

Toothpaste, bleach, cleaning agents, all are acidic in nature.

0%
True
0%
False

Statement A: Tamarind contains tartaric acid
Statement B- Vinegar contains acetic acid
Choose the correct option:

0%
Statement A is correct and Statement B is incorrect.
0%
Both Statement A and Statement B are incorrect.
0%
Statement A is incorrect and Statement B is correct.
0%
Both Statement A and B are correct.

Explanation

Tamarind pulp contains tartaric acid and vinegar contains acetic acid. Therefore, option

$D$ is correct.

Litmus and turmeric are synthetic indicators.

0%
True
0%
False

Sodium hydroxide is acidic in nature.

0%
True
0%
False

Which of the following mixtures in aqueous solution of equimolar concentration acts as a buffer solution?

0%
$H NO_{3} + Na OH$
0%
$H_{2}SO_{4} + K OH$
0%
$NH_{4}OH(excess) + H Cl$
0%
$CH_{3}COO H + Na OH (excess)$

Explanation

A mixture of ammonium hydroxide and

$HCl$ react to form ammonium chloride. This also contains unreacted ammonium hydroxide.

$NH_4OH+HCl \rightarrow NH_4Cl + H_2O$

Thus, the resulting mixture of

$(NH_4OH+NH_4Cl)$ is a basic buffer solution. It contains a mixture of weak base ammonium hydroxide and its salt (ammonium chloride) with strong acid

$(HCl)$ .

It is used in a qualitative analysis of group

$III$ radicals.

A mixture of

$HNO_3$ and

$NaOH$ is a mixture of strong acid and strong base similarly a mixture of

$H_2SO_4$ and

$KOH$ is also a mixture of strong acid and strong base, thus they do not form buffer solution.

Similarly, in option D mixture will form a strong base and salt of strong base weak acid hence it will not form a buffer solution.

Hence, option C is correct.

$HCl$ does not behave as acid in __

0%
Water
0%
Ammonia
0%
Benzene
0%
Aqueous NaOH

Explanation

Acids in aprotic solvents lose their acidic character due to decrease in their dissociation due to aprotic nature of solvents.

Because benzene is a non-polar solvent.

$HCl$ will dissolve in it, but won’t ionize into

$Cl^-$ and

$H^+$ ions.

Option C is correct.

$H_{2}O$ is a:

0%
Proton acceptor
0%
Proton donor
0%
Both proton donor and proton acceptor
0%
Neither proton donor, nor proton acceptor

Explanation

Hint:

According to Bronsted-Lowry's theory, the compounds that have tendency to donate $\left(\mathrm{H}^{+}\right)$ ion are acids and the compounds that have tendency to accept $\left(\mathrm{H}^{+}\right)$ ion are bases.

Explanation:

A compound such as water

$\left(\mathrm{H}_{2} \mathrm{O}\right)$ has many interesting properties. Water molecules can accept a proton to act as Bronsted-Lowry bases in certain circumstances. The following is an example of

$\mathrm{HCl}$ dissolving in water:

$\mathrm{HCl}+\mathrm{H}_{2} \mathrm{O}_{(\ell)} \rightarrow \mathrm{H}_{3} \mathrm{O}_{\text {(aq) }}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}$

The another possibility is water can act like a Bronsted-Lowry acid by donating a proton. Water donates a proton to a proton-accepting amide ion in the presence of ammonia, resulting in the following product:

$\mathrm{H}_{2} \mathrm{O}_{(\ell)}+\mathrm{NH}_{2(a q)}^{-} \rightarrow \mathrm{OH}_{(a q)}^{-}+N H_{3(a q)}$

Therefore, depending on the circumstances,

$\left(\mathrm{H}_{2} \mathrm{O}\right)$ can either act as a Bronsted-Lowry acid or a Bronsted-Lowry base. There are other substances that can react either as acids or bases in certain circumstances, but water is the most common example. It is called an amphiprotic compound when it either donates or accepts a proton, depending on the situation.

Final Answer: Option (C)

In aqueous solution, the following mixture acts as buffer :

0%
$H NO_{3} + KN O_{3}$
0%
$H_{2}SO_{4} + K_{2}SO_{4}$
0%
$NH_{4}OH + NH_{4}Cl$
0%
$CH_{3}COO H+ Na Cl$

Explanation

For buffer solution
Weak acid + Its salt of strong base
or
Weak base + Its salt of strong acid.

In Option C,

$NH_{4}OH \to$ weak base

$NH_{4}Cl \to$ weak base and strong acid salt

Thus option C is right. While no other option follow definition of buffer solution.

The mixed solution of pthallic acid and potassium hydrogen pthallate is _________.

0%
Basic buffer
0%
Acid buffer
0%
Not a buffer
0%
An acid

The ionic product of water changes when:

0%
an acid is added to it
0%
a base is added to it
0%
either a base or an acid is added to it
0%
the temperature is changed

Explanation

Hint: $\text { Ionization of } \mathrm{H}_{2} \mathrm{O} \text { increases as the temperature increases. }$

Correct Answer:

$D$

Explanation:

Water is a weak electrolyte and ionizes only to a small extent.

water is dissociated in the following manner:

$\Rightarrow \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-}$

According to Le-Chatelier’s principle, as the temperature increases, the position of equilibrium shifts to the right of the equation to minimize the effect of increased temperature.

This means with the increase in temperature the concentration of

$\left[\mathrm{H}^{+}\right]$ and

$\left[\mathrm{OH}^{-}\right]$ ions will increase which will result in increased value ofK

$_{w}$ as forward reaction is favoured.

Conclusion:

This means that the ionic product depends only on temperature and is unaffected by any other changes.

Which can act as buffer?

0%
$NH_{4}Cl + H Cl$
0%
$CH_{3}COO H+H_{2}CO_{3}$
0%
40 ml of 0.1 M $NaCN$ +20 ml of 0.1 M $HCN$
0%
$NaCl$ + $NaOH$

Explanation

Buffer solution is a mixture of weak acid and its salt of weak acid.

In Option C

$NaCN\rightarrow$ salt of strong base and weak acid

$HCN \rightarrow$ weak acid

$\therefore$ They can act as buffer.

While no other options follow definition of buffer solution.

Option C is correct.

Acid buffer can be prepared by mixing solutions of:

0%
$Na Cl + Na OH$
0%
$H_{2}SO_{4} + Na_{2}SO_{4}$
0%
$NH_{4}OH + NH_{4}Cl$
0%
$CH_{3} COO Na + CH_{3}COO H$

Explanation

a solution of a weak acid + salt of the weakacid and a strong base; is called an acidic buffer and here mixture of

$CH_3COONa$ and

$CH_3COOH$ is a mixture of weak acid and its salt of strong base,

What will happen when

$CH_{3}COO Na$ is added to an aqueous solution of

$CH_{3}COO H$ ?

0%
the pH of the solution decreases
0%
the pH of the solution increases
0%
the pH of the solution remains unaltered
0%
an acidic salt is produced

Explanation

$CH_{3}COOH \leftrightharpoons CH_{3} COO^{-}+H^{+}$

$CH_{3}COONa$ is added, equilibrium will shift backward due to the common ion effect.

$H^{+}$ will decrease.

$\therefore$ pH will increase.

Which of the following statements is true about buffer solution?

0%
It keeps the pH value constant in a chemical reaction.
0%
It decreases the pH value in a chemical reaction.
0%
It increases the pH value in a chemical reaction.
0%
It first increases and then decreases the pH value in a chemical reaction.

Which of the following is a buffer solution?

0%
$Na OH +CH_{3}COO Na$
0%
$Na OH+Na_{2}SO_{4}$
0%
$K_{2}SO_{4} + H_{2}SO_{4}$
0%
$NH_{4}OH+NH_{4}Cl$

Explanation

Hint- A buffer solution is prepared by mixing an equal amount of weak base and its salt.

Correct option : D

Explanation of correct option:

A buffer solution is a solution which only changes slightly when adding an acid or base to it. It consists of a weak acid and its conjugate base for an acid-buffer solution. It consists of a weak base and its conjugate acid for a basic buffer solution.

Among all given options

$NH_{4}OH+NH_{4}Cl$ is a combination of weak base

$NH_{4}OH$ and its salt

$NH_{4}Cl$ , hence, it will act as a buffer solution.

$NH_4OH+HCl \rightarrow NH_4Cl + H_2O$

Thus, the resulting mixture

$(NH_4OH+NH_4Cl)$ is a basic buffer.

Explanation of incorrect options:

$A$ is a combination of a strong base and weak acid salt , hence it will not form a buffer solution.
$B$ is a combination of a strong base and strong acid-base salt , hence it will not form a buffer solution.
$C$ is a combination of strong acid-base salt , hence it will not form a buffer solution.

A solution which maintains constant pH when small amounts of acid or alkali are added is known as ____.

0%
Indicator
0%
Buffer
0%
Amphoteric
0%
Neutral

The solution that is made by mixing appropriate amounts of a weak acid and its salt of strong base is called ________.

0%
acidic buffer
0%
basic buffer
0%
neutral buffer
0%
universal indicator

The solutions which tend to keep the concentration of hydrogen ions constant, even when small amounts of strong acid or strong base are added to them, are known as:

0%
isohydric solutions
0%
buffer solutions
0%
isotonic solutions
0%
neutral solutions

Match List-1 with List-2.

List -1	List-2
A) Aq. Solution of NaCl	1)Acidic buffer
B) Aq. Solution of $CH_3COONa$	2)Acidic solution
C) Aq. Solution of $NH_{4}Cl$	3) Basic solution
D) Aq. Solution of $CH_{3}COOH$ along with $CH_{3}COONa$	4)Neutral solution

0%
$A-1; B-2 ; C-3; D-4$
0%
$A-4 ; B-3; C-2; D-1$
0%
$A-3; B- 4; C-2; D-1$
0%
$A-4; B-3; C-1; C-2$

Explanation

$NaCl$ is a salt of strong acid

$HCl$ and strong base

$NaOH$ . It does not undergo hydrolysis in water. Hence its aqueous solution is neutral.

Sodium acetate is a salt of strong base sodium hydroxide and weak acid acetic acid. Its solution is basic in nature.

Ammonium chloride is the salt of strong acid

$HCl$ and weak base ammonium hydroxide. Its solution is acidic in nature.

A mixture of acetic acid and sodium acetate is an acidic buffer. Its aqueous solution is acidic in nature.

Thus, A-4 ; B-3; C-2; D-1

Hence, the correct option is

$\text{B}$

Which of the following solution cannot act as a buffer?

0%
$Na H_{2}PO_{4} + H_{3}PO_{4}$
0%
$CH_{3}COO H+CH_{3}COO Na$
0%
$H Cl + NH_{4}Cl$
0%
$H_{3}PO_{4} + Na_{2} H PO_{4}$

Explanation

A mixture of weak acid and a salt of strong base, or a mixture of weak base and a salt of strong acid acts as a buffer.

$NaH_2PO_4+H_3PO_4$ ,

$CH_3COOH+CH_3COONa$ and

$H_3PO_4+Na_2HPO_4$ have an acid/salt-base pair.

Therefore they act as a buffer.

A mixture of strong acid and its salt of strong acid don't make an effective buffer.

Therefore,

$HCl+NH_4Cl$ cannot act as a buffer.

Hence the correct option is C.

Which of the following mixtures in aqueous solution acts as a buffer?

0%
$H NO_{3} + K NO_{3}$
0%
$H_{2}SO_{4} + K_{2}SO_{4}$
0%
$NH_{4}OH + NH_{4}Cl$
0%
$CH_{3}COO H + Na Cl$

Explanation

A mixture of ammonium hydroxide and

$HCl$ react to form ammonium chloride. This also contains unreacted ammonium hydroxide.

$NH_4OH+HCl \rightarrow NH_4Cl + H_2O$

Thus, the resulting mixture of

$(NH_4OH+NH_4Cl)$ is a basic buffer solution. It contains a mixture of weak base ammonium hydroxide and its salt (ammonium chloride) with a strong acid (

$HCl$ ).

Option A and B contain strong acid and salt of strong acid and strong base thus it does not form buffer solution,

Option D contains a mixture of weak acid and salt of strong acid and strong base thus it also does not form a buffer solution.

Buffer solution can be obtained by mixing aqueous solution of _______.

0%
Sodium acetate and excess of HCl
0%
Sodium acetate and acetic acid
0%
Sodium chloride and HCl
0%
Acetic acid and excess of NaOH

Which of the following solutions can act as buffer?

0%
$0.1$ molar aq. $NaCl$
0%
$0.1$ molar aq. $CH_{3}COOH + 0.1$ molar $NaOH$
0%
$0.1$ molar aq. ammonium acetate
0%
$0.1$ molar $H_{3}PO_{4}$

Explanation

Weak acid and salts of a weak acid and weak base and salt of weak base can act as a buffer in the aqueous medium.

0.1 molar aq. ammonium acetate solution is salt of weak base ammonium hydroxide and weak base is ammonium hydroxide.

Thus it forms a buffer solution.

Whereas 0.1 molar aq. NaCl is a salt of strong acid HCl and strong base NaOH, 0.1 molar aq. acetic acid and 0.1 molar NaOH contain strong base NaOH completely neutralised the weak acid. Thus it does not form buffer solution.

And 0.1 molar

$H_3PO_4$ is a strong acid thus it also does not form buffer solution.

Option C is correct.

Which of the following will suppress the ionization of acetic acid in aqueous solution?

0%
$NaCl$
0%
$HCl$
0%
$KCl$
0%
Unpredictable

Explanation

Common ion effect is observed when a solution of weak electrolyte is mixed with a solution of a strong electrolyte, which provides an ion common to that provided by the weak electrolyte.

Acetic acid is a weak electrolyte.

$HCl$ is a strong electrolyte.

$HCl$ provides hydrogen ion which is common to that provided by acetic acid.

Option B is correct.

Purification of

$NaCl$ by the passage of

$HCl$ through brine solution is based on which of the following?

0%
Distribution coefficient
0%
Le Chatelier’s principle
0%
Displacement Law
0%
Common ion effect

Explanation

Purification of

$NaCl$ by the passage of

$HCl$ through brine is based on the common ion effect.

$HCl$ is a strong electrolyte and provides an ion

$(Cl^-)$ that is common to that provided by the weak electrolyte. Thus, the ionization of weak electrolytes is suppressed.

$80^{\circ} C$ , distilled water has hydronium ion

$\left (H_{3}O^{+} \right )$ concentration equal to

$1\times 10^{-6} mol/l$ . The value of

$K_w$ at this temperature would be:

0%
$1\times 10^{-6}$
0%
$1\times 10^{-12}$
0%
$1\times 10^{-9}$
0%
$1\times 10^{-14}$

Explanation

The distilled water is neutral with equal concentrations of hydrogen ion and hydroxide ion.

Hence,

$K_w=[H^+][OH^-]$

$={1 \times 10^{-6}} \times {1 \times 10^{-6}}$

$=1 \times 10^{-12}$ .

Hence, the correct option is

$\text{B}$

Which of the following pairs will show common ion effect?

0%
$NaC N + K C N$
0%
$Na Cl + H Cl$
0%
$NH_{4} OH + NH_{4} Cl$
0%
$Ba Cl_{2}+ Ba\left ( NO_{3} \right )_{2}$

Explanation

Common ion effect is observed when a solution of weak electrolyte is mixed with a solution of strong electrolyte which provides an ion common to that provided by a weak electrolyte.

Ammonium hydroxide is a weak electrolyte and ammonium chloride is a strong electrolyte. Ammonium chloride provides ammonium ions common to ammonium hydroxide.

Thus, the pair

$NH_{4} OH + NH_{4} Cl$ shows common ion effect.

All other pairs have a mixture of strong electrolytes mixed with strong electrolyte hence they do not show common ion effect.

$90^{\circ}C$ for pure water,

$[H_{3}O^{+}] =10^{-6.5}$ mol/litre. The value of

$K_{w}$ at this temperature is:

0%
$10^{-6}$
0%
$10^{-13}$
0%
$10^{-14}$
0%
$10^{-8}$

Explanation

Pure water is neutral having equal concentrations of hydrogen ion and hydroxide ion.

$K_w$ is a self-ionization constant which is obtained from a self-ionization reaction, as given below:

$H_2O + H_2O \Longleftrightarrow H_3O^+ + OH^{-}$ where

$[H_3O^+] =[OH^-] = 1 \times 10^{-6.5}$

$K_w=[H_3O^+][OH^-]$

Hence,

$K_w ={1 \times 10^{-6.5}} \times {1 \times 10^{-6.5}}=1 \times 10^{-13}$ .

Option B is correct.

In the third group of qualitative analysis, the precipitating reagent is

$NH_{4}Cl + NH_{4} OH$ . The function of

$NH_{4}Cl$ is to_______

0%
increase the ionization of $NH_{4} OH$
0%
suppress the ionization of $NH_{4} OH$
0%
stabilise the hydroxides of group cations
0%
convert the ions of group third into their respective chlorides

Explanation

Common ion effect is observed when a solution of weak electrolyte is mixed with a solution of a strong electrolyte, which provides an ion common to that provided by a weak electrolyte.

The

$NH_4OH$ is weak base it does not ionises completely. Thus due to presence of common ion

$NH_4^+$ in

$NH_4Cl$ , it suppresses the ionisation of weak base

$NH_4OH$ in order to decrease the

$OH^-$ concentration so that higher group cations will not get precipitated.

Thus the pair

$NH_{4} OH + NH_{4} Cl$ shows common ion effect.

Ammonium chloride suppresses the ionization of ammonium hydroxide.

Option B is correct.

Buffer solution is prepared by mixing:

0%
Strong acid + its salt of strong base
0%
Weak acid + its salt of strong base.
0%
Strong acid + its salt of weak base
0%
None of the above

The addition of

$HCl$ will not supress the ionization of ________.

0%
$CH_{3}COOH$
0%
$H_{2}S$
0%
$C_6H_5COOH$
0%
$H_{2}SO_{4}$

Explanation

Common ion effect is observed when a solution of weak electrolyte is mixed with a solution of strong electrolyte, which provides an ion, common to that provided by weak electrolyte.

$H_{2}SO_{4}$ and

$HCl$ are strong electrolytes. Hence, addition of

$HCl$ will not suppress the ionization of

$H_{2}SO_{4}$ .

And if

$HCl$ is added to any other solution which are given ,then it will suppress ionization because all other solutions are strong electrolyte.

The strength of acid is highest in:

0%
$p K_{a}=6$
0%
$p K_{a}=5$
0%
$p K_{a}=10$
0%
$p K_{a}= 1$

Explanation

$pK_a = -log\:K_a$

Higher the

$K_a$ , higher is the strength of the acid.

For higher

$K_a$ ,

$pK_a$ value is smaller.

Option D is correct.

5.6 grams of KOH

$(M. wt = 56)$ is present in 1 litre of solution. Its pH is:

0%
1
0%
13
0%
14
0%
0

Explanation

Molarity of KOH solution =

$=\cfrac {\text{Weight}} {\text{Molecular weight} \times \text{volume}}= \cfrac {5.6} {56 \times 1}=0.1$ M.

This is equal to

$[OH^-]$ .

$pOH=-log[OH^-]=-log( 0.1 )=1$

$pH=14-pOH=14-1=13$ .

Hence, option B is correct.

The number of

$H^{+}$ ions in

$1\ cc$ of a solution of

$pH=13$ is:

0%
$6.023\times 10^{7}$
0%
$1\times 10^{-13}$
0%
$6.023\times 10^{13}$
0%
$1\times 10^{16}$

Explanation

pH 13 means

$[H^+]=10^{-13}$ M.
Thus, 1 L of solution contains

$10^{-13}$ moles of hydrogen ions.
Hence, 1 ml or 1 cc of solution will contain

$10^{-13} \times \dfrac {1} {1000}=10^{-16}$ moles of hydrogen ions.
They corresponds to

$10^{-16} \times 6.023 \times 10^{23}=6.023 \times 10^{7}$ hydrogen ions.

The

$pH$ of a mono-acidic base is 12.The molarity of the base is:

0%
0.02 moles/litre
0%
0.05 moles/litre
0%
0.5 moles/litre
0%
0.2 moles/litre

Explanation

$pH+pOH=14$ .

Thus,

$pOH=14-pH=14-12.6690=1.3310$ .

But,

$pOH=-log[OH^-]$ .

Hence,

$[OH^-]=0.05\ M$ .

This is equal to the molarity of base.

Option B is correct.

Identify the acid which can give acidic salt and base which can give basic salt.

0%
$H_{2} SO _{4} , \ Ca(OH)_{2}$
0%
$Ba(OH)_{2}, \ HCl$
0%
$H_{2} CO_{3} , \ Al(OH)_{3}$
0%
$CH_{3}COOH, \ Mg(OH)_{2}$

Explanation

$H_{2} SO_{4} + NaOH \rightarrow NaHSO_{4} + H_{2} O$ Acidic salt.

$H_{2} CO_{3} + NaOH \rightarrow NaHCO_{3} + H_{2} O$
Similarly bases

$Ca(OH)_{2}, Al(OH)_{3}$ can also forms basic salts.

8 gram of NaOH is mixed with 9.8 gram of

$H_{2}SO_{4}$ , the pH of the solution is:

0%
more than 7
0%
7
0%
less than 7
0%
cant be said

Explanation

8 gram (0.2 mole) of NaOH (molecular weight 40 g/mol) completely neutralizes 9.8 gram (0.1 mole) of

$H_2SO_4$ (molecular weight 98 g/mol).

$NaOH\to Na^++OH^-$

$H_2SO_4\to 2H^++SO_4^{2-}$

Since the molar concentration of both

$OH^-$ and

$H^+$ are same, the resulting solution will be neutral. Its pH will be 7.

The solubility of

$AgCl$ in

$NaCl$ solution is less than that in pure water, because of the ________.

0%
solubility product of $AgCl$ is less than of $NaCl$
0%
common ion effect
0%
both $A$ and $B$
0%
none of these

Explanation

Since,

$NaCl$ is soluble to a very significant extent, when

$AgCl$ is added to

$NaCl$ solution, the common ion

$[Cl^-]$ increases in the solution. To have the solubility product or

$K_{sp}$ of

$AgCl$ constant,

$[Ag^+]$ will decrease or

$AgCl$ will percipitate out from the solution. This is common ion effect. Hence solubility of

$AgCl$ in

$NaCl$ solution will be less than that in pure water.

A weak mono acidic base is

$0.5$ % ionized in

$0.01 M$ solution. The hydroxide ion concentration in the solution is:

0%
$5 \times 10^{-2}$
0%
$5 \times 10^{-5}$
0%
$5 \times 10^{-10}$
0%
$2 \times 10^{-11}$

Explanation

Let the monoacidic base be

$BOH$ .

Given that

$c=0.01M$ and

$\alpha =0.005$ .

SInce

$\alpha\ll 1$

On dissociation,

$BOH\leftrightharpoons B^{+}+OH^{-}$

$c$

$-$

$c(1-\alpha)$

$c\alpha$

The concentration of

$OH^{-}$ is given by

$c\alpha$ .

$\Rightarrow \left [ OH^{-} \right ]=0.01\times 0.005$

$\Rightarrow \left [ OH^{-} \right ]=5 \times 10^{-5}$ .

Hence, option B is correct.

$pH$ of a solution produced when aqueous solution of

$pH=6$ is mixed with an equal volume of an aqueous solution of

$pH=3$ is about:

0%
$4.3$
0%
$3.3$
0%
$4.0$
0%
$4.5$

Explanation

The

$pH$ of solution is negative logarithm of hydrogen ion concentration.

$pH=-log[H^+]$

Hence,

$[H^+]=10^{-pH}$

$pH=3$ and

$pH=6$ corresponds to hydrogen ion concentrations of

$0.001 M$ and

$0.000001 M$ respectively.

Assuming

$1 L$ of each solution is mixed,
the concentration of resulting solution is

$\dfrac{0.001 M + 0.000001 M}{2} = 0.0005M$ .

$pH = -log [H^+] = -log\left( 0.0005\right)=3.3$

Option B is correct.

When ammonium chloride is added to ammonia solution, the pH of the resulting solution will be:

0%
increased
0%
seven
0%
decreased
0%
not changed

Explanation

Common ion effect is observed when a solution of weak electrolyte is mixed with a solution of strong electrolyte which provides an ion common to that provided by a weak electrolyte.

Ammonium hydroxide is a weak electrolyte and ammonium chloride is a strong electrolyte. Ammonium chloride provides ammonium ion which is common to that provided by ammonium hydroxide. Thus, the pair

$NH_{4} OH + NH_{4} Cl$ shows common ion effect. Ammonium chloride suppresses the ionization of ammonium hydroxide.

This decreases the hydroxide ion concentration and increases the hydrogen ion concentration. Hence, the pH value decreases.

Amongst the following hydroxides, the one which has the highest value of

$K_{sp}$ at

$25^{\circ}$

$C$ is:

0%
$KOH$
0%
$CsOH$
0%
$LiOH$
0%
$RbOH$

Explanation

Going down Group-I in the periodic nature, the ionic nature of the hydroxides increases. As ionic nature increases, the solubility also increases. Hence, the solubility product is maximum for

$CsOH$ .

The

$pH$ of a solution obtained by mixing

$50\: ml$ of

$0.4N$

$HCl$ and

$50\: ml$ of

$0.2 N$

$NaOH$ is:

0%
$- log\ 2$
0%
$-log\ (0.2)$
0%
$1.0$
0%
$2.0$

Explanation

The expression for the hydrogen ion concentration is as given below.

$[H^+]= \dfrac {V_aN_a-V_bN_b} {V_a+V_b}$

Substitute values in the above expression,

$[H^+]= \dfrac {50 \times 0.4-50 \times0.2} {50+50}=0.1M$

$pH=-log\ [H^+]=-log\ (0.1)=1$

Solubility of

$AgCl$ will be minimum in ___________.

0%
0.01 M $Na_{2} SO_{4}$
0%
0.01 M $Ca Cl_{2}$
0%
Pure water
0%
0.001 M $Ag NO_{3}$

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 2 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 2 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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