Very Short Answer type
Question 1
(a)Write the $K_{SP}$ expression of the zirconium phosphate $({\rm Zr}^{4+})_3({\rm PO}_4^{3-})_4$.
(b) Calculate the pH of ${10}^{-8}$ M HCl.
(c)which of the following salt will give highest pH in water
(i) NaCl
(ii) KCl
(iii)$CuSO_4$
(iv) $Na_2CO_3$
(d) Write the solubility product expression for the following salts: Silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide.
(e) which of the following is the buffer solution?
(i)HCl and KCl
(ii) $NaNO_2$, $HNO_2$
(iii) NaOH and HCl
(iv) $NH_4NO_3$, $HNO_3$
(f)What is the pH Scale?
(g)What is lewis Acid and base?
(h)The solubility of a lead iodide ${\rm Pbl}_2$ is S mol/L. Calculate its $K_{sp}$.
(i)The solubility of magnesium hydroxide is 0.0093 mol/L, calculate its $K_{sp}$.
(j)The solubility of ${\rm CaF}_2$ in water at 298 K is $1.7 \times 10^{-3}$ grams per $100cm^3$. Calculate the solubility product of ${\rm CaF}_2$ at 298 K.
(k)The Ksp of lead fluoride is $2.7 \times 10^{-8}$. Calculate the molar solubility and solubility in g/L of ${\rm PbF}_2$.
(l)Determine the solubility of barium sulphate in 0.05 M in Barium chloride solution. The $K_{sp}$ of $BaSO_4=1.0 \times 10^{-10}$.
(m)What is the minimum amount of water required to dissolve 1g of calcium sulphate at 298K? $K_{sp}\left(CaSO_4\right)=9.1 \times{10}^{-6}$.
(n)Which of the following is not a Lewis Acid
(i) $BF_3$
(ii)$FeCL_3$
(iii) $SiF_4$
(iv) $C_2H_4$
Solution
a. $K_{sp} = 6912 (S)^7$
b.pH=8
c.$Na_2CO_3$
e. $NaNO_2$, $HNO_2$
n. $C_2H_4$
Short answer type
Question 2
What is a common-ion effect?
Solution
It can be defined as a shift in equilibrium on adding a substance that provides more of an ionic species already present in the dissociation equilibrium. Thus, we can say that common ion effect is a phenomenon based oil the Le Chatellier's principle
Question 3
Give two uses of common-ion effect.
Solution
(i) It is used in the complete precipitation of ${\rm Al}^{3+}$ ion in the Group III of the qualitative analysis.
(ii) It is also used in obtaining high purity NaCl by bubbling HCl gas through it.
Question 4
What is Henderson-Hasselbalch equation?
Solution
This is an equation relating the pH of a buffer for different concentrations of conjugate acid and base.
For an acidic buffer, $pH=pK_a+log([salt]/[acid])$.
For a basic buffer, $pOH=pK_b+log([salt]/[base])$.
Question 5
Calculate the molar solubility of $Ni(OH)_2$ in 10 M NaoH. The ionic product of $Ni(OH)_2$ is $2.0 \times 10^{-15}$
Solution
Let the solubility of Ni(OH)2 be equal to S.
Dissolution of S mol/L of Ni(OH)2 provides S mol/L of Ni2+ and 2S mol/L of OH−, but the total concentration of OH− = (0.10 + 2S) mol/L because the solution already contains 0.10 mol/L of OH− from NaOH.
$K_{sp}= 2.0 \times 10^{−15}$
$K_{sp} = [Ni2+][OH−]^2$
$2.0 \times 10^{−15} = (S)(0.10 + 2s)^2 $
Ksp is small, 2S << 0.10
∴
Therefore
$2 \times 10^{−15} = (S) (0.10)^2$
$S= 2 \times 10^{−13}$
Question 6
The values of Ksp of two sparingly soluble salts $Ni(OH)_2$ and AgCN are $2.0 \times 10^{-15}$ and $6 \times 10^{-17}$ respectively. Which salt is more soluble? Explain.
Solution
$AgCN \rightleftharpoons Ag+ + CN–$
$K_{sp} = [Ag+][CN–] = 6 \times 10^{–17}$
$Ni(OH)_2 \rightleftharpoons Ni2+ + 2OH–$
$K_{sp} = [Ni2+][OH–]^2 = 2 \times 10^{–15}$
Let [Ag+] = S1, then [CN-] = S1
Let [Ni2+] = S2, then [OH–] = 2S2
$S_1^2 = 6 \times 10^{–17}$ , $S_1 = 7.8 \times 10^{–9}$
$(S_2)(2S_2)^2 = 2 \times 10^{–15}$, $S_2 = 0.58 \times 10^{–4}$
Therefore Ni(OH)2 is more soluble than AgCN
Question 7
The solubility product constant of ${\rm Ag}_2CrO_4$ and AgBr are $5.0 \times 10^{-13}$ respectively. Calculate and ratio of the.molarities of their saturated solution.
Question 8
Will AgCl be more soluble in pur water or aqueous NaCl solution and why?
Solution
In NaCl aqueous. solution the Cl-ions will increases. Since solubility products, $Ksp=[Ag^+][Cl^-]$ remains constant. $[{\rm Ag}^+]$ will decrease. Therefore, solubility of AgCl will be less in NaCl solution than in water i.e. according to Le-Chatelier's principle reaction will move towards backward direction.
Question 9
The pKa of acetic acid and pKb of ammonium hydroxide are 7.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution.
Solution
$pH=7+1/2(pK_a-pK_b)=7+1/2[4.76-4.75]=7+1/2[0.01]=7+0.005=7.005$
Question 10
Predict if the solutions of the following salts are neutral, acidic of basic:
NaCl,KBr, NaCN,${\rm NH}_4{\rm NO}_3$,$NaNO_2$ and KF
Solution
(NaCl-> neutral), (KBr -> neutral), (NaCN- >basic), ${\rm NH}_4{\rm NO}_3$ ->acidic, ${\rm NaNO}_2$ -> basic),KF ->basic.
Question 11
What will be the decreasing order of basic strength of the following conjugate bases?
${\rm OH}^-$,${\rm RO}^-$,${\rm CH}_2C{\rm OO}^-$,${\rm Cl}^-$.
Solution
$RO^-$ ->$OH^-$ -> $CH_2COO^-$ ->$Cl^-$
Question 12
Arrange the following in increasing order of pH.
${\rm KNO}_3\left(aq\right)$,${\rm CH}_3COONa\left(aq\right)$,${\rm NH}_4Cl\left(aq\right)$,$C_6H_5COONH_4\left(aq\right)$.
Solution
${\rm NH}_4Cl$ < $C_6H_5COONH_4$ < ${\rm KNO}_3$ < ${\rm CH}_3COONa$
Question 13
Why is it essential to add dil. HCl before proceeding to test for the metals for the II group?
Solution
H2S is a week acid and ionizes as follows :
$H_2S \rightleftharpoons 2H+ + S^{2-}$
HCL ionizes as follows :
$HCL \rightleftharpoons H+ + Cl^{-}$
The addition of HCL which is highly ionizes increased the concentration of H+ ions ,so the concentration of $S^{2-}$ decrease due to common ion effect
Hence the sulphides of the II group metals get precipitated as there solubility product is small
Since the solubility products for the surface of groups 3rd and 4th cations is very high those cations are not precipitate out under the above conditions.
Numericals
Question 14
Calculate [H+] when
(i) pH = 6.83
(ii) pH = 7.38
(iii) pH = 6.4
Solution
(i)
pH= - log [H+]
6.83 = - log [H+]
Anti log – 6.83 = [H+]
- 6.83 + 7 – 7 = [H+]
-7 + 0.17 = [H+]
[H+]= $1.479 \times 10^{-7}$
(ii) pH = - log [H+]
7.38 = - log [H+]
-7.38 = log [H+]
Anti log – 7.38 = [H+]
-7.38 + 8 – 8 = [H+]
-8 + 0.62 = [H+]
[H+]=$ 4.169 \times 10^{-8}$
(iii)
pH = - log [H+]
6.4 = - log [H+]
-6.4 = log [H+]
Anti log – 6.4 = [H+]
-6.4 + 7 – 7 = [H+]
0.6 – 7 = [H+]
[H+]=$7.782 \times 10^{-7}$
Question 15
Equal volumes of 0.002 M solutions of sodium chloride and silver nitrate are mixed together. Will its lead to precipitation of $AgCl(K_{sp})=10^{-10}$?
Question 16
Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH=6 and pH=4 respectively.
Question 17
The solubility product of $Al(OH)_3$ is $2.7 \times {10}^{-11}$. Calculate its solubility in g/L and also find out pH of this solution (Atomic mass of Al=27 u).
Question 18
Which of the following will produce a buffer solution when mixed in equal volumes?
(i) 0.1 $mol/dm^3$ ${\rm NH}_4OH$ and 0.1 $mol/dm^3$ HCl
(ii) 0.05 $mol/dm^3$ ${\rm NH}_4OH$ and 0.1 $mol/dm^3$ HCl
(iii) 0.1 $mol/dm^3$ ${\rm NH}_4OH$ and 0.05 $mol/dm^3$ HCl
(iv) 0.1 $mol/dm^3$ ${\rm CH}_4COONa$ and 0.1 $mol/dm^3$ NaOH
Question 19
Two gram of TiOH dissolved in water to give 2 litre soln. 204 = Ti; O = 16 gm; H = 1 gm. Give pH of the solution.
Solution
Molecular weight of TiOH = 204 + 16 + 1
= 221 gm
Molarity = $\frac{Weight\ }{Molar\ mass}\ \times\ \frac{1}{vol.^\prime l^\prime}$
$= \frac{2}{221}\ \times\frac{1}{2}$
$= \frac{1}{221}=0.004524$
$= 4.5 \times 10^{-3}$
= 360.6551
= 2.4
pKw = pH + pOH
14 =m pH + 2.4
14 – 2.4 = pH
11.6 = pH
Question 20
0.3 g of Ca(OH)2 gives 500 ml of sol. Give pH of the solution.
Solution
Molecular mass of Ca(OH)2 = 40 + (16 + 1) x 2
= 40 + 34
= 74 gm
Multiply with '2' as OH are 2
Molarity = $\frac{Weight}{Molecular\ mass}\ \times\frac{1000}{500}\ \times 2$
$= \frac{0.3}{74}\ \times2\ \times2=\frac{0.3}{37}\ \times 2$
$=6 \div 370 = 0.0162$
$[OH] = 16.2 \times 10^{-3}$
$pOH = -log [OH-]$
$= - [log (16.2 \times 10^{-3}]
=1.791
pH=14 -1791=12.21
Question 21
$FeO (s) + CO (g) \rightleftharpoons FeO + O_2 (g)$
Kp = 0.265 atm at 1050 K
Initial pressure : pCO2 = 0.8, pCO = 1.4
Find the equilbrium pressure of CO & CO2
Solution
$Q = \frac{\left[CO_2\right]}{\left[CO\right]} = \frac{0.8}{1.4}=0.57$
since, Kp < Q
Therefore Reaction goes backward and reaction is not at equilibrium.
As reaction goes backward pCO2 will decrease & pCO will increase.
Let charge in pressure be ‘p’
Increase in pressure of CO = 1.4 + p
Decrease in pressure of CO2 = 0.8 – p
$Kp = \frac{PCO_2}{pCO}=>0.265=\ \frac{0.8-p}{1.4+p}$
(0.265) (1.4 + p) = 0.8 – p
p = 0.338
∴ pCO = 1.4 + p = 1.4 + 0339 = 1.739
pCO2 = 0.8 – p = 0.461
Question 22
In the following reaction if initially 10 moles of PCl5 are taken which are 5% dissociated after reaching the equilibrium. Give the value of Equilibrium constant.
Solution
$PCl_5 \rightleftharpoons PCl_3 + Cl_2$
1 0 0
$\frac{0.05}{10(1-0.05)}$ $\frac{0.05}{0.05}$ $\frac{0.05}{0.05}$
$Kc = \frac {[PCl3 [Cl2]}{[PCl5]}$
$K_c = \frac{0.05\ \times 0.05}{10-0.5}$
$= 2.63 \times 10^{-4}$
Question 23
$2 NOCl (g) \rightleftharpoons 2NO (g) + Cl2 (g)$
$K_p = 1.8 \times\ {10}^{-2}$ at 500 K
Solution
$K_p = KC (RT)^ {\Delta n}$
$\Delta n=3-2=1$
$K_p = K_c (0.0831 \times 500 )$
$K_c =\frac{1.8\ \times{10}^{-2}\ \times\ {10}^{-2}\ }{.831}=2.16 \times 10^{-4}$
Question 24
The value of Kp for the reaction,
$CO2 (g) + C (s) \rightleftharpoons 2CO (g)$
is 3.0 at 1000 K. If initially pCO2 = 0.48 bar and PCO = 0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO2.
Solution
For the reaction,
let ‘x’ be the decrease in pressure of CO2,
then
$CO2 (g) + C (s) \rightleftharpoons 2CO (g)$
Initial pressure: 0.48 bar 0
At equilibrium:
(0.48 – x)bar 2x bar
$K_p = \frac {(pCO)^2}{pCO2}$
$K_p = \frac {(2x)^2}{(0.48 – x)} = 3$
$4x^2 = 3(0.48 – x)$
Solving this and ignoring the negative value
x = 0.33
The equilibrium partial pressures are,
pCO2= 2x = 2 × 0.33 = 0.66 bar
pCO2= 0.48 – x = 0.48 – 0.33 = 0.15 bar
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