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CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 14 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Equilibrium
Quiz 14
At $$90^{ \circ }C$$, pure water has $$[{ H }_{ 3 }O^{ + }]={ 10 }^{ -6.7 }\ mol/lit$$. What is the value of $${ K }_{ W }$$ at $${ 90 }^{ \circ }C :-$$
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$${ 10 }^{ -6 }$$
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$${ 10 }^{ -12 }$$
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$${ 10 }^{ -14 }$$
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$${ 10 }^{ -13.4 }$$
Which of the following will not function as buffer solution?
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$${\text{Na}}{{\text{H}}_{\text{2}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{ + N}}{{\text{a}}_{\text{2}}}{\text{HP}}{{\text{O}}_{\text{4}}}\left( {{\text{1:1}}\,{\text{Molar}}\,{\text{Ratio}}} \right)$$
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$${\text{N}}{{\text{H}}_4}{\text{Cl + NaOH}}\left( {{\text{2:1}}\,{\text{Molar}}\,{\text{Ratio}}} \right)$$
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$${\text{HCOOH + NaOH}}\left( {{\text{1:1}}\,{\text{Molar}}\,{\text{Ratio}}} \right)$$
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$${\text{HCOOH + HCOONa}}\left( {{\text{1:1}}\,{\text{Molar}}\,{\text{Ratio}}} \right)$$
In which of the following solution of electrolyte, ionic equilibrium is set up?
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$$H_{2}SO_{4}$$
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$$HCl$$
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$$NaOH$$
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$$HCN$$
Which of the following is a buffer solution?
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500 mL of 0.1 N CH$$_3$$COOH +500 mL of 0.4 N NaOH
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500 mL of 0.1 N CH$$_3$$COOH +500 mL of 0.1 N HCl
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500 mL of 0.1 N CH$$_3$$COOH +500 mL of 0.2 N NaOH
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500 mL of 0.1 N CH$$_3$$COOH +500 mL of 0.3N NaOH
On adding NaCl solution to the saturated solution of $$PbCl_2$$ then the concentration of :
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$$Pb^{2+}$$ will increase
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$$Cl^-$$ will decrease
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$$Pb^{2+}$$ will decrease
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Ions will remain unaltered
The ionic product of water at $$ 25^oC $$ is $$ 10^{-14} $$ The ionic product at $$ 90^oC $$ will be
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$$ 1 \times 10^{-20} $$
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$$ 1 \times 10^{-12} $$
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$$ 1 \times 10^{-14} $$
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$$ 1 \times 10^{-16} $$
In a buffer solution the ratio of concentration of $${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\,\,{\text{and}}\,{\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}\,\,{\text{is}}\,\,{\text{1:1}}$$ when it changes in 2:1 what will be the value of pH of buffer:-
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0%
Increase
0%
Decrease
0%
No effects
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N. O. T.
A hydrogen electrode placed in a buffer solution of NaCl and HCN in the ratio of X:Y and Y:X has an oxidation electrode potential values $${E_1}\,\,and\,\,{E_2}$$ volts respectively at $${25^0}C$$. The pKa value of HCN is
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$$\frac{{({E_1} + {E_2})}}{{0.118}}$$
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$$ - \frac{{({E_1} + {E_2})}}{{0.118}}$$
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$$\frac{{({E_1} - {E_2})}}{{0.118}}$$
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$$\frac{{({E_2} - {E_1})}}{{0.118}}$$
If the degree of ionization of water is 1.8 x $$10^{-9}$$ at a given temperature. Its ionic product will be
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1.8 x $$10^{-16}$$
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1 x $$10^{-14}$$
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1 x $$10^{-16}$$
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1.67 x $$10^{-14}$$
$$2{H_2}O\leftrightharpoons {H_3}{O^ + } + O{H^ - },{K_w} = 1\times {10^{ - 14}}$$ at $${25^ \circ }C$$ hence $$K_a$$ is :
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$$1\times 10^{-14}$$
0%
$$18\times 10^{-17}$$
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$$5.55\times 10^{-3}$$
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$$1.00\times 10^{-7}$$
The dissociation constant of a weak acid is $$1.0 \times { 10 }^{ -5 }$$, the equilibrium constant for its reaction with strong base is
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0%
$$1.0 \times { 10 }^{ -5 }$$
0%
$$1.0 \times { 10 }^{ 9 }$$
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$$1.0 \times { 10 }^{ 7 }$$
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$$1.0 \times { 10 }^{ 14 }$$
One litre of water contains $$10 ^ { - 7 }$$ mole of $$H ^ { + }$$ ions. Degree of ionisation of water is:
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$$1.8 \times 10 ^ { - 7 } \%$$
0%
$$0.8 \times 10 ^ { - 9 } \%$$
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$$3.6 \times 10 ^ { - 9 } \%$$
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$$3.6 \times 10 ^ { - 7 } \%$$
Explanation
$$1$$ litre of water contains $$1000/18$$ mole.
So degree of ionization $$=\dfrac {10^{-7}\times 18}{1000}=1.8\times 10^{-7}\%$$
Option A is correct.
Aqueous solution of NaCl is neutral because
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$$Na^{ + }$$ undergoes hydrolysis
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Cl undergoes hydrolysis
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Both $$Na^{ + }$$ and Cl undergo hydrolusis
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Does not undergo hydrolysis
Which of the following may be regarded as buffer solution?
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50 ml of 0.1 M acetic acid + 50 ml of 0.05 M $$NaOH$$
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50 ml of 0.05 ammonia + 50 ml of 0.1 M $$HCl$$
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25 ml of 0.1 M acetic acid + 50 ml of 0.1 M $$NaOH$$
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50 ml of 0.1 ammonia + 50 ml of 0.05 M $$HCl$$
The total number of different kind of buffer obtained during the titration of $${ H }_{ 3 }{ PO }_{ 4 }$$ with NaOH are:-
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3
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1
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2
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0
Select the correct acid-base equilibrium.
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$$\mathrm { HNO } _ { 3 } + \mathrm { HF } \rightleftharpoons \mathrm { H } _ { 2 } \mathrm { NO } _ { 3 } ^ { + } + \mathrm { F } ^ { - }$$
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$$\mathrm { HNO } _ { 3 } + \mathrm { HF } \rightleftharpoons \mathrm { H } _ { 2 } \mathrm { F } ^ { + } + \mathrm { NO } _ { 3 } ^ { - }$$
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$$\mathrm { CH } _ { 3 } \mathrm { COOH } + \mathrm { NH } _ { 3 } \rightleftharpoons \mathrm { CH } _ { 3 } \mathrm { COOH } _ { 2 } ^ { + } + \mathrm { NH } _ { 2 } ^ { + }$$
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$$\mathrm { HF } + \mathrm { H } _ { 2 } \mathrm { O } \rightleftharpoons \mathrm { H } _ { 2 } \mathrm { F } ^ { + } + \mathrm { OH } ^ { - }$$
$$HCOOH$$ and $$CH_3COOH$$ solutions have equal $$pH$$. If $$\cfrac{K_1}{K_2}$$ (ratio of ionisation constants of acids) is 4, their molar concentration ratio will be:
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0%
2
0%
0.5
0%
4
0%
0.25
At $$90 ^ { \circ } C ,$$ pure water has $$\left[ \mathrm { H } _ { 3 } \mathrm { O } ^ { + } \right] = 10 ^ { - 6 } \mathrm { mol }$$ litre $$^ { - 1 } .$$ The value of $$K _ { w }$$ at $$90 ^ { \circ } \mathrm { C }$$ is:
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0%
$$10 ^ { - 6 }$$
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$$10 ^ { - 12 }$$
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$$10 ^ { - 14 }$$
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$$10 ^ { - 8 }$$
Hydrogen ion concentration of an aqueous solution is $$1 \times 10^{-4}M$$. The solution is diluted with equal volume of water. Hydroxyl ion concentration of the resultant solution in terms of mol $$dm^{-3}$$ is __________.
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$$1 \times 10^{-8}$$
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$$1 \times 10^{-6}$$
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$$2 \times 10^{-10}$$
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$$0.5 \times 10^{-10}$$
Explanation
$$\text{Concentration of aqueous solution=} 10^{−4}M$$
$$\text{If we add equal volume, let say V of water. The final volume of solution will be 2V. The new concentration is}$$
$$⇒M_1V_1=M_2V_2$$
$$⇒10^{−4}M×V=M_2×2V$$
$$⇒M_2=5×10−5M ⟶ \text{New concentration of }[H^+] .$$
$$pH=−log[H^+]=−log[5×10^{−5}]$$
$$⇒pH=4.3$$
$$⇒pOH=14−pH=9.69$$
$$⇒−log[OH^-]=9.69$$
$$⇒[OH^-]=10^{−9.69}=2×10^{−10}$$
$$\text{Thus the concentration of }[OH^-] \ is\ 2×10^{−10}M .$$
If $$1M$$ $$C{H_3}COONa$$ is added to $$1M$$ $$C{H_3}COOH$$, then:
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$$pH$$ of the solution increases
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$$pH$$ decreases
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$$pH$$ does not change
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cannot be predicted
The dissociation constants for acetic acid and HCN at $$25^{0}$$C are $$1 . 5 \times 10^{-5}$$ and $$4 . 5 \times 10 ^{-10} $$, respectively. The equilibrium constant for the equilibrium $$CN^{-} + CH_{3}COOH \leftrightharpoons HCN + CH_{3}COO^{-}$$ would be :
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0%
$$3\times10^{4}$$
0%
$$3\times10^{5}$$
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$$3\times10^{-5}$$
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$$3\times10^{-4}$$
At $${ 25 }^{ \circ }C$$ the dissociation constant of a base , BOH, is $$1.0\times { 10 }^{ -12 }$$. The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be
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0%
$$1.0\times { 10 }^{ -6 }{ molL }^{ -1 }$$
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$$1.0\times { 10 }^{ -7 }{ molL }^{ -1 }$$
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$$2.0\times { 10 }^{ -6 }{ molL }^{ -1 }$$
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$$1.0\times { 10 }^{ -5 }{ molL }^{ -1 }$$
A solution is saturated with respect to $$AgSCN$$ as well as $$AgBr$$. The concentration of $$Ag^{+}$$ in the solution would be:
($$ K_{sp}$$ for $$AgSCN = 1.2 \times 10^{-12}$$ and for $$AgBr = 5 \times 10^{-13}$$)
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$$1.3 \times 10^{-5}$$
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$$1.3 \times 10^{-12}$$
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$$1.3 \times 10^{-6}$$
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$$1.3 \times 10^{-8}$$
Which of the following salt evolve(s) $${ N }_{ 2 }$$ gas on heating ?
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$$\left( { NH }_{ 4 } \right) _{ { 2 } }{ CrO }_{ 4 }$$
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$${ NH }_{ 4 }{ NO }_{ 2 }$$
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$${ NH }_{ 4 }{ NO }_{ 3 }$$
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$$\left( { NH }_{ 4 } \right) _{ { 2 } }{ CO }_{ 3 }$$
Which of following salt solution will impart greenish colour ?
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$$ K_2Cr_2O_7 $$
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$$ ZnCl_2 $$
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$$ K_2MnO_4 $$
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$$ K_2CrO_4 $$
10 ml of 0.2 M $$ AgNO_3 $$ is mixed with 10 ml of 0.1 M $$NaCl$$. What is the concenctration of $$ Cl^- $$ in the resulting solution $$[ K_{sp} (AgCl) = 10^{-10} M^2] $$
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$$ 10^{-5} M $$
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$$ 10^{-9} M $$
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$$ 2 \times 10^{-9} M $$
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$$ 10^{-8} M $$
For preparing a buffer solution of pH = 7.0, which buffer system you will choose?
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$$H_3PO_4,H_2PO_4^-$$
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$$H_2PO_4^-, HPO_4^{2-}$$
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$$HPO^{2-}_4, PO^{3-}_4$$
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$$H_3PO_4, PO_4^{3-}$$
The percentage of pyridine $$\left( { C }_{ 5 }{ H }_{ 5 }N \right) $$ that forms pyridinium ion $$\left( { C }_{ 5 }{ H }_{ 5 }{ N }^{ + }H \right) $$ in a 0.10 m aqueous pyridine solution $$\left( K_{ b } \ for\ { C }_{ 5 }{ H }_{ 5 }N=1.7\times { 10 }^{ -9 } \right) $$ is:
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0%
0.0060%
0%
0.013%
0%
0.77%
0%
1.6%
A weak acid react with strong base, ionisation constant of weak acid is $${ 10 }^{ -4 }$$. Find out equilibrium constant for this reaction.
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$${ 10 }^{ -10 }$$
0%
$${ 10 }^{ 10 }$$
0%
$${ 10 }^{ -9}$$
0%
$${ 10 }^{ 9}$$
Which statement is/are correct-
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If $$0.1\ M$$ $$ CH_3COOH $$ aqueous solution $$ ( K_a = 1.8 \times 10^{-5} ) $$ is diluted at $$ 2560^o C $$, then $$ [H^+] $$ will increase
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On heating a sample of pure water $$ [H^+] $$ and $$ [OH^-] $$ increases but the sample still remains neutral
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$$ 10^{-4}\ M$$ $$HCl$$ solution is less acidic than $$0.1\ M$$ $$HCN$$ $$ ( K_a HCN = 10^{-5} ) $$
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$$ 10^{-8}\ M $$ $$HCl$$ aqueous solution will have pH exactly equal to 8
The self ionisation constant of $$\mathrm { NH } _ { 3 }$$ at $$50 ^ { \circ } \mathrm { C }$$ is given by $$\mathrm { K } _ { \mathrm { NH } _ { 3 } } = \left[ \mathrm { NH } _ { 4 } ^ { + } \right] \left[ \mathrm { NH } _ { 2 } ^ { - } \right] = 10 ^ { - 30 }$$. How many $$N H _ { 2 } ^ { - }$$ ions are present per $$\mathrm { cm } ^ { 3 }$$ of pure liquid $$\mathrm { NH } _ { 3 }$$?
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$$6 \times 10 ^ { 6 }$$
0%
$$6 \times 10 ^ { 5 }$$
0%
$$6 \times 10 ^ { - 5 }$$
0%
$$6 \times 10 ^ { - 6 }$$
The dissociation constant of an acid HA is $$1 \times 10 ^ { - 5 }$$. The pH of 0.1 molar solution of the acid will be approximately:
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0%
3
0%
5
0%
1
0%
6
The silicate anion in the mineral kinetic is a chain of three $$SiO_4^{4^-}$$ tetrahedral that shares conners with adjacent tetrahedral. The mineral also contains $$Ca$$ icon $$Cu$$ ions and water molecules in a 1:1:1 ratio mineral represented as:
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$$CaCu_2Si_{ 3 }O_{ 10 }.H_{ 2 }O$$
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$$CaCuSi_{ 3 }O_{ 10 }.H_{ 2 }O$$
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$$Ca_{ 2 }Cu_{ 2 }Si_{ 3 }O_{ 10 }.2H_{ 2 }O$$
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None of these
Which of the following is ionic?
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$$HCl$$
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$$CHCl_3$$
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$$IF_5$$
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$$KI$$
For which salt the $$\mathrm { pH }$$ of its solution does not change with dilution?
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$$\mathrm { NH } _ { 4 } \mathrm { Cl }$$
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$$\mathrm { CH } _ { 3 } \mathrm { COONH } _ { 4 }$$
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$$\mathrm { CH } _ { 3 } \mathrm { COONa }$$
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None of these
Consider the following equilibrium.
Liquid $$\rightleftharpoons$$ Vapour
Which of the following relation is correct?
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$$\dfrac{d ln P}{dT}=\dfrac{\Delta Hv}{RT^2}$$
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$$\dfrac{d ln G}{dT^2}=\dfrac{\Delta Hv}{RT^2}$$
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$$\dfrac{d ln P}{dT}=\dfrac{-\Delta Hv}{RT}$$
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$$\dfrac{d ln P}{dT^2}=\dfrac{-\Delta Hv}{T^2}$$
The $$pH$$ of an aqueous solution of sodium chloride at $$60^{\circ}C$$ is
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$$7.0$$
0%
$$> 7.0$$
0%
$$<7.0$$
0%
$$0$$
Explanation
We know that at $$25^{\circ}C$$ dissociation constant $$K_w$$ of water is,
$$H_2O\rightleftharpoons H^++OH^-$$
$$[H^+][OH^-]=10^{-14}=K_w$$
So, $$[H^+]=[OH^-]=10^{-7}M$$
Thus $$pH=7$$ for $$H_2O$$ at
$$25^{\circ}C$$
Adding $$NaCl$$ does not affect the $$pH$$ whatever be the temperature because sodium chloride is salt of strong acid and strong base. So, it is $$H^+$$ ion of water which is affecting the $$pH$$.
Dissociation of water is an endothermic process. With an increase in
temperature
dissociation
constant of water increases. So the dissociation constant is more i.e., $$K_w>10^{-14}$$.
Therefore, $$[H^+]>10^{-7}$$ and $$pH$$ is less than 7.
Hence the correct option is (C).
The EMF of cell: $$ H_{2}(g)|$$ Buffer $$ \parallel $$ Normal calmelelectrod, is 0.70 V at $$ 25^{\circ}C,$$ when barometric pressure is 760 mm. What is the pH of the buffer solution? $$ E_{Calomel}^{0} = 0.28 V. [2.303 RTIF = 0.06]$$
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0%
3.5
0%
7.0
0%
tending to zero
0%
tending to 14.0
The dissociation constant of formic acid is $$0.00024$$. The hydrogen ion concentration in $$0.002 M \ HCOOH$$ solution is nearly
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$$6.93 \times 10^{-4}M$$
0%
$$4.8 \times 10^{-7}M$$
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$$5.8 \times 10^{-4}M$$
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$$1.4 \times 10^{-4}M$$
The degree of dissociation of pure water at $$25^{\circ}C$$ is found to be $$1.8 \times 10^{-9}$$. The dissociation constant, $$K_{d}$$ of water, at $$25^{\circ}C$$ is
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0%
$$10^{-14}$$
0%
$$1.8 \times 10^{-16}$$
0%
$$5.56 \times 10^{-13}$$
0%
$$1.8 \times 10^{-14}$$
Explanation
Given:
Degree of dissociation of pure water, $$\alpha =1.8\times 10^{-9}$$
We know that,
Density of pure water = $$1\space g\space cm^{-3}=1000\space g\space L^{-1}$$
Molar mass of water = $$18\space g\space mol^{-1}$$
$$Concentration=\frac{number\space of\space moles}{Volume}=\frac{1000\space g\space L^{-1}}{18\space g\space mol^{-1}}=55.5\space M$$
Dissociation of water, $$H_2O\rightleftharpoons H^++OH^-$$
Dissociation constant, $$K_d=\frac{[H^+][OH^-]}{[H_2O]}=\frac{C\alpha\times C\alpha}{C-C\alpha}$$
As $$\alpha<<<1$$, $$[H_2O]$$ is not affected.
Therefore,
$$K_d=c\alpha^2=55.5\times (1.8\times 10^{-9})^2=1.79\times 10^{-16}\approx 1.8\times 10^{-16}$$
Hence, the correct option is (B).
Which one of the following salts gives aqueous solution which is neutral?
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$$NaHC{ O }_{ 3 }$$
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$$NaHS{ O }_{ 4 }$$
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$$NaCl$$
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$$\\ Na{ H }_{ 4 }HC{ O }_{ 3 }$$
When ammonia is added to pure water, the concentration of which of the following species already present, will decrease?
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0%
$$OH^-$$
0%
$$H_3O^+$$
0%
$$NH_4^+$$
0%
None of these
n-caproic acid, $$C_{5}H_{11}COOH$$, found in coconut and palm oil is used in making artificial flavors, has solubility in water equal to $$11.6 g/L$$. The saturated solution has $$pH = 3.0$$. The $$K_{a }$$ of acid is
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0%
$$10^{-6}$$
0%
$$10^{-5}$$
0%
$$2 \times 10^{-5}$$
0%
$$2 \times 10^{-6}$$
What is $$ [H^+] $$ in mol/L of a solution that is 0.20 M $$ CH_3CCONa $$ and 0.10 M in $$ CH_3COOH $$?
$$( K_a\ for\ CH_3COOH = 1.8 \times 10^{-5} ) $$
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0%
$$ 3.5 \times 10^{-4} $$
0%
$$ 1.1 \times 10^{-5} $$
0%
$$ 1.8 \times 10^{-5} $$
0%
$$ 9.0 \times 10^{-6} $$
A solution has initially $$0.1 M -HCOOH$$ and $$0.2 M- HCN$$. $$K_{a}$$ of $$HCOOH = 2.56 \times 10^{-4}$$, $$K_{a}$$ of $$HCN = 9.6 \times 10^{-10}$$. The only incorrect statement for the solution is $$(log 2 = 0.3)$$
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$$[H]^{+} = 1.6 \times 10^{-3} M$$
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$$[HCOO]^{-} = 1.6 \times 10^{-3} M$$
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$$[CN]^{-} = 1.2 \times 10^{-7} M$$
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$$pOH = 2.8$$
Calculate the formation constant for the reaction of a tripositive metal ion with thiocyanate ions to form the monocomplex if the total metal concentration in the solution is $$2 \times 10^{-3} M$$, the total $$SCN^{-}$$ concentration is $$1.51 \times 10^{-3} M$$ and the free $$SCN^{-}$$ concentration is $$1.0 \times 10^{-5} M$$.
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0%
$$7.55 \times 10^{4}$$
0%
$$3 \times 10^{5}$$
0%
$$3.33 \times 10^{-6}$$
0%
$$1.5 \times 10^{5}$$
Small amount of freshly precipitated magnesium hydroxides are stirred vigorously in a buffer solution containing $$0.25 M$$ of $$NH_{4}Cl$$ and $$0.05 M$$ of $$NH_{4}OH$$. $$[Mg^{2+}]$$ in the resulting solution is ($$K_{b}$$ for $$NH_{4}OH = 2.0 \times 10^{-5}$$ and $$K_{sp}$$ of $$Mg(OH)_{2} = 8.0 \times 10^{-12}$$)
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0%
$$4 \times 10^{-6} M$$
0%
$$2 \times 10^{-6} M$$
0%
$$0.5 M$$
0%
$$2.0 M$$
A volume of $$1.0 L$$ of solution which was in equilibrium with solid mixture of $$AgCl$$ and $$Ag_{2}CrO_{4}$$ was found to contain $$1 \times 10^{-4}$$ moles of $$Ag^{+}$$ ions, $$1.0 \times 10^{-6}$$ moles of $$CI^{-}$$ ions and $$8.0 \times 10^{-4}$$ moles of $$CrO_{4}^{2-}$$ ions. $$Ag^{+}$$ ions are added slowly to the above mixture (keeping the volume constant) till $$8.0 \times 10^{-7}$$ moles of $$AgCl$$ got precipitated. How many moles of $$Ag_{2}CrO_{4}$$ were precipitated simultaneously?
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0%
$$7.68 \times 10^{-4}$$
0%
$$4.8 \times 10^{-4}$$
0%
$$8.0 \times 10^{-4}$$
0%
$$7.68 \times 10^{-5}$$
Which one of the following is not an acid salt?
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0%
$$NaH_2PO_2$$
0%
$$NaH_2PO_3$$
0%
$$NaH_2PO_4$$
0%
$$Na_2HPO_4$$
Explanation
$$NaH_2PO_2$$ is not an acidic salt. It is a basic salt. It is a salt of weak acid $$H_3PO_2$$ and a strong base $$NaOH$$. Hence, the aqueous solution of $$NaH_2PO_2$$ is alkaline in nature. In $$NaH_2PO_2,$$ there is no acidic hydrogen.
Hence, the correct option is (A).
The solubility of $$AgCN$$ in a buffer solution of $$pH - 3.0$$ is ($$K_{sp}$$ of $$AgCN = 1.2 \times 10^{-16}$$; $$K_{a}$$ of $$HCN = 4.8 \times 10^{-10}$$)
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0%
$$1.58 \times 10^{-5}M$$
0%
$$2.0 \times 10^{-5}M$$
0%
$$1.58 \times 10^{-4}M$$
0%
$$2.5 \times 10^{-9}M$$
Explanation
Assume solubility of AgCN be a M in presence of a buffer of pH = 3
Given :- $$K_{sp}$$ of AgCN = $$1.2 \times 10^{-16}$$ and $$ K_{a} $$ for HCN = $$4.8 \times 10^{-10}$$
$$ AgCN \rightleftharpoons Ag^{+} + CN^{-}$$
$$ [Ag^{+}] = a; \quad [H^+] = 10^-{3}M; $$
For weak HCN, $$ K_{a} = \frac{[H^{+}][CN^{-}]}{[HCN]} $$
$$\Rightarrow 4.8 \times 10^{-10} = \dfrac{[10^{-3}][CN^{-}]}{a} $$
$$ \therefore [CN^{-}] = 4.8 \times 10^{-7} \times a $$
$$ For \ AgCN : K_{sp} = Ag^{+} \times CN^{-} $$
$$ 1.2 \times 10^{-16} = a \times 4.8 \times 10^{-7} \times a $$
$$ a^{2} = \dfrac{1.2 \times 10^{-16}}{4.8 \times 10^{-7}} $$
$$ a = 1.58 \times 10^{-5} M $$
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