a. $K_{sp} = 6912 (S)^7$
b.pH=8
c.$Na_2CO_3$
e. $NaNO_2$, $HNO_2$
n. $C_2H_4$
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
(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.
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, $pOH=pK_b+log([salt]/[base])$.
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
$2 \times 10^{−15} = (S) (0.10)^2$
$S= 2 \times 10^{−13}$
$AgCN \rightleftharpoons Ag+ + CN–$
$K_{sp} = [Ag+][CN–] = 6 \times 10^{–17}$
$Ni(OH)_2 \rightleftharpoons Ni2+ + 2OH–$
$K_{sp} = [Ni2+][OH–]^2 = 2 \times 10^{–15}$
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
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.
$pH=7+1/2(pK_a-pK_b)=7+1/2[4.76-4.75]=7+1/2[0.01]=7+0.005=7.005$
(NaCl-> neutral), (KBr -> neutral), (NaCN- >basic), ${\rm NH}_4{\rm NO}_3$ ->acidic, ${\rm NaNO}_2$ -> basic),KF ->basic.
$RO^-$ ->$OH^-$ -> $CH_2COO^-$ ->$Cl^-$
${\rm NH}_4Cl$ < $C_6H_5COONH_4$ < ${\rm KNO}_3$ < ${\rm CH}_3COONa$
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.
(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+]=$7.782 \times 10^{-7}$
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
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
$Q = \frac{\left[CO_2\right]}{\left[CO\right]} = \frac{0.8}{1.4}=0.57$
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
$PCl_5 \rightleftharpoons PCl_3 + Cl_2$
1 0 0
$\frac{0.05}{10(1-0.05)}$ $\frac{0.05}{0.05}$ $\frac{0.05}{0.05}$
$Kc = \frac {[PCl3 [Cl2]}{[PCl5]}$
$K_c = \frac{0.05\ \times 0.05}{10-0.5}$
$= 2.63 \times 10^{-4}$
$K_p = KC (RT)^ {\Delta n}$
$\Delta n=3-2=1$
$K_p = K_c (0.0831 \times 500 )$
$K_c =\frac{1.8\ \times{10}^{-2}\ \times\ {10}^{-2}\ }{.831}=2.16 \times 10^{-4}$
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
