The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium nitrate with 100 mL of 0.10 M sodium iodate is $\boldsymbol{X} \times 10^{-6} \mathrm{~mol} \mathrm{dm}^{-3}$. The value of $\boldsymbol{X}$ is ____________.
Use: Solubility product constant $\left(K_{\mathrm{sp}}\right)$ of barium iodate $=1.58 \times 10^{-9}$
Explanation:
Calculate the concentration of iodate ions $\left[\text{IO}_3^-\right]$ in the equilibrium mixture:
The concentration in the prepared solution is given by:
$ \left[\text{IO}_3^-\right]_{\text{eqm}} = \frac{6}{300} = 0.02 \, \text{M} $
Write the expression for the solubility product constant ($ K_{\text{sp}} $) for barium iodate:
$ K_{\text{sp}} = [\text{Ba}^{2+}][\text{IO}_3^-]^2 $
Rearrange to solve for the concentration of barium ions $[\text{Ba}^{2+}]$:
$ [\text{Ba}^{2+}] = \frac{1.58 \times 10^{-9}}{(0.02)^2} = 3.95 \times 10^{-6} \, \text{M} $
The solubility of $\text{Ba}(\text{IO}_3)_2$, which is the same as the concentration of barium ions in this context, is:
$ X = 3.95 $
Thus, the solubility constant $ X $ is $ 3.95 $, and the solubility of barium iodate is $ 3.95 \times 10^{-6} \, \text{mol dm}^{-3} $.
At 25 °C, the concentration of H+ ions in 1.00 × 10−3 M aqueous solution of a weak monobasic acid having acid dissociation constant (Ka) of 4.00 × 10−11 is X × 10−7 M. The value of X is ______.
Use: Ionic product of water (Kw) = 1.00 × 10−14 at 25 °C
Explanation:
Concentration of $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{Na}_{2} \mathrm{SO}_{4}$ in a solution is $1 \mathrm{M}$ and $1.8 \times 10^{-2} \mathrm{M}$, respectively. Molar solubility of $\mathrm{PbSO}_{4}$ in the same solution is $\mathrm{X} \times 10^{-\mathrm{Y}} \mathrm{M}$ (expressed in scientific notation). The value of $Y$ is ________.
[Given: Solubility product of $\mathrm{PbSO}_{4}\left(K_{s p}\right)=1.6 \times 10^{-8}$. For $\mathrm{H}_{2} \mathrm{SO}_{4}, K_{a l}$ is very large and $\left.K_{a 2}=1.2 \times 10^{-2}\right]$
Explanation:
Conentration of $\mathrm{H}_2 \mathrm{SO}_4=1 \mathrm{M}$
Conentration of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ = $1.8 \times 10^{-2} \mathrm{M}$
From the above two equations, we get
$ \left[\mathrm{H}^{+}\right]=1 \mathrm{M} \text { and }\left[\mathrm{SO}_4^{2-}\right]=1.8 \times 10^{-2} $
$ K_c=\frac{1.8 \times 10^{-2} \times 1}{1}=1.8 \times 10^{-2} $
and it is given that $\mathrm{K}_{a_2}\left(\mathrm{Q}_c\right)=1.2 \times 10^{-2} \mathrm{M}$
Since, $\mathrm{K}_{a_2}$ (i.e., $\mathrm{Q}_c$ ) $<\mathrm{K}_c$
$\therefore$ Rather than dissociation of $\mathrm{HSO}_4^{-}$into $\mathrm{H}^{+}$and $\mathrm{SO}_4^{2-}$ ions, association between already present $\mathrm{H}^{+}$and $\mathrm{SO}_4^{2-}$ will take place.
Assuming ' $x$ ' mol/L of $\mathrm{SO}_4^{2-}$ and $\mathrm{H}^{+}$combines to form $\mathrm{HSO}_4^{-}$
$\begin{aligned} & \mathrm{K}_{a 2}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{SO}_4^{2-}\right]}{\left[\mathrm{HSO}_4^{-}\right]} \\\\ & 1.2 \times 10^{-2}=\frac{(1-x)\left(1.8 \times 10^{-2}-x\right)}{(1+x)} \\\\ & \because x<<1, \text { so }(1+x) \approx 1 \text { and }(1-x) \approx 1 \\\\ & 1.2 \times 10^{-2}=1.8 \times 10^{-2}-x \\\\ & x=\left(1.8 \times 10^{-2}\right)-\left(1.2 \times 10^{-2}\right) \\\\ & x=0.6 \times 10^{-2} \mathrm{M} \\\\ & \text { So, }\left[\mathrm{SO}_4^{2-}\right]=1.8 \times 10^{-2}-x \\\\ & {\left[\mathrm{SO}_4^{2-}\right] }=\left(1.8 \times 10^{-2}\right)-\left(0.6 \times 10^{-2}\right) \\\\ & {\left[\mathrm{SO}_4^{2-}\right] }=1.2 \times 10^{-2} \mathrm{M}\end{aligned}$
Given, $\quad \mathrm{K}_{s p}=1.6 \times 10^{-8}$
$ \therefore $ $y\left(1.2 \times 10^{-2}+y\right)=1.6 \times 10^{-8}$
Since, $y<<1$, So $1.2 \times 10^{-2}+y \approx 1.2 \times 10^{-2}$
So, $y \times 1.2 \times 10^{-2}=1.6 \times 10^{-8}$
$ \Rightarrow $ $y =\frac{1.6 \times 10^{-8}}{1.2 \times 10^{-2}}$
$ \Rightarrow $ $ y =1.33 \times 10^{-6} $
$ X \times 10^{-Y} \mathrm{M} =1.33 \times 10^{-6} \mathrm{M} $
So, $\mathrm{Y}=6$
Hence, the value of $Y$ is 6 .
A solution is prepared by mixing $0.01 \mathrm{~mol}$ each of $\mathrm{H}_{2} \mathrm{CO}_{3}, \mathrm{NaHCO}_{3}, \mathrm{Na}_{2} \mathrm{CO}_{3}$, and $\mathrm{NaOH}$ in $100 \mathrm{~mL}$ of water. $p \mathrm{H}$ of the resulting solution is _________.
[Given: $p \mathrm{~K}_{\mathrm{a} 1}$ and $p \mathrm{~K}_{\mathrm{a} 2}$ of $\mathrm{H}_{2} \mathrm{CO}_{3}$ are $6.37$ and 10.32, respectively; $\log 2=0.30$ ]
Explanation:
First acid base reaction between H2CO3 and NaOH takes place.
In the final solution, we have 0.01 mole Na2CO3 and 0.02 moles of NaHCO3.
Here, we have a buffer of NaHCO3 and Na2CO3.
$\therefore$ $pH = p{K_{{a_2}}} + \log {{[Salt]} \over {[Acid]}}$
$ = 10.32 + \log {{\left( {{{0.01} \over {0.1}}} \right)} \over {\left( {{{0.02} \over {0.1}}} \right)}}$
$ = 10.32 + \log {1 \over 2}$
$ = 10.32 - \log 2$
$ = 10.32 - 0.3$
$ = 10.02$
$\therefore$ $pH = 10.02$
$B + HA\buildrel {} \over \longrightarrow B{H^ + } + {A^ - }$
Explanation:
$pOH = p{K_b} + \log {{[B{H^ + }]} \over {[B]}}$
At half equivalence point :
[BH+] = [B] ($ \because $ pH = 11)
Therefore, pOH = pKb = 14 $-$ 11 = 3
$ \because $ pKb = 3.00
Use Ksp(ZnS) = 1.25 $ \times $ 10$-$22 and overall dissociation constant of
H2S, Knet = K1K2 = 1 $ \times $ 10-21.
Explanation:
${K_{sp}}(ZnS) = [Z{n^{2 + }}][{S^{2 - }}] = 1.25 \times {10^{ - 22}}$
$0.05 \times [{S^{2 - }}] = 1.25 \times {10^{ - 22}}$
$ \Rightarrow [{S^{2 - }}] = {{1.25 \times {{10}^{ - 22}}} \over {0.05}} \Rightarrow 25 \times {10^{ - 22}}M$
For H2S, $\mathop {{H_2}S}\limits_{(0.1M)} \buildrel {} \over \longrightarrow 2{H^ + } + \mathop {{S^{ - 2}}}\limits_{(25 \times {{10}^{ - 22}}M)} $
${K_{net}} = 1 \times {10^{ - 21}} = {{{{[{H^ + }]}^2}[{S^{2 - }}]} \over {[{H_2}S]}}$
$1 \times {10^{ - 21}} = {{{{[{H^ + }]}^2} \times 25 \times {{10}^{ - 22}}} \over {[0.1]}}$
${[{H^ + }]^2} = {1 \over {25}}$
$[{H^ + }] = {1 \over 5} = 0.2M$
(Given that the value of solubility product of $AB$ $\left( {{K_{sp}}} \right) = 2 \times {10^{ - 10}}$ and the value of ionization constant of $HB$ $\left( {{K_a}} \right) = 1 \times {10^{ - 8}}$)
Explanation:
Let solubility of AB in the buffer of pH 3 = x
K3 = ${{{K_{sp}}} \over {{K_a}}}$
${K_3} = {{[HB][{A^ + }]} \over {[{H^ + }]}} = {{{K_{sp}}} \over {{K_a}}}$
$\therefore$ ${{{x^2}} \over {({{10}^{ - 3}})}} = {{2 \times {{10}^{ - 10}}} \over {1 \times {{10}^{ - 8}}}}$
$\therefore$ x = 4.47 $\times$ 10$-$3 M = y $\times$ 10$-$3 M
$\therefore$ y = 4.47
Explanation:
(i) The solubility of $\mathrm{AgCl}(\mathrm{s})$ in saturated solution is expressed as :
$ \mathrm{H}_2 \mathrm{O}+\mathrm{AgCl}_{(s)} \rightarrow \underset{x}{\mathrm{Ag}^{+}}(a q)+\underset{x} {\mathrm{Cl}^{-}}(a q) $
Though $\mathrm{AgCl}(\mathrm{s})$ has low solubility with $\mathrm{K}_{s p}=1.6$ $\times 10^{-10}$ still some silver $\left(\mathrm{Ag}^{+}\right)$and chloride $\left(\mathrm{Cl}^{-}\right)$ are dissolved in solution. Let the concentration of these ions be $x \mathrm{~mol}^{-1}$ (in solution).
$ \begin{aligned} {[\mathrm{Ag}]^{+} } & =\left[\mathrm{Cl}^{-}\right]=x \\ & =1.6 \times 10^{-10} \\ \mathrm{CuCl}(s)+\mathrm{H}_2 \mathrm{O}(l) & \rightleftharpoons \mathrm{Cu}^{+}(a q)+\mathrm{Cl}^{-}(a q) \end{aligned} $
(ii) Similarly though CuCl has low solubility in aqueous solution $\left[K_{s p}=1.0 \times 10^{-6}\right]$, still some copper $\left(\mathrm{Cu}^{+}\right)$and chloride $\left(\mathrm{Cl}^{-}\right)$are dissolved in the solution. Let the concentration of these ions be $y \mathrm{~mol} \mathrm{~L}^{-1}$ (in solution).
$ \left[\mathrm{Cu}^{+}\right]=\left[\mathrm{Cl}^{-}\right]=y $
(iii) The salts $\mathrm{AgCl}(\mathrm{s})$ and $\mathrm{CuCl}(\mathrm{s})$ are present in equilibrium with their ions as follows :
$\text{AgCl}(s) + \text{H}_2\text{O} \rightleftharpoons \underset{x}{\text{Ag}^+ (aq)} + \underset{x+y}{\text{Cl}^- (aq)}$
$K_{sp}(\text{AgCl}) = 1.6 \times 10^{-10} = x(x + y) \quad \ldots \text{(i)}$
$\text{CuCl}_{(s)} + \text{H}_2\text{O} \rightleftharpoons \underset{x}{\text{Cu}^+ (aq)} + \underset{y+x}{\text{Cl}^- (aq)}$
$K_{sp}(\text{CuCl}) = 1.0 \times 10^{-6} = y \times (x + y) \quad \ldots \text{(ii)}$
Dividing equation (ii) by (i) :
$ \frac{1.0 \times 10^{-6}}{1.6 \times 10^{-10}} = \frac{y}{x} $
$ \frac{y}{x} = \frac{1.0}{1.6} \times 10^4 $ …(iii)
No. of moles of CuCl = $0.1 \text{ mol}$
Volume of solution = $1 \text{ L}$
$\begin{aligned} & \text { Concentration of } \mathrm{CuCl}=\frac{\text { No. of moles of } \mathrm{CuCl}}{\text { Volume of solution }} \\\\ & \text { Concentration of } \mathrm{CuCl}=\frac{0.1 \mathrm{~mol}}{1 \mathrm{~L}}=0.1\end{aligned}$
$\begin{aligned} \mathrm{K}_{s p} & =1.0 \times 10^{-6}=y \times y \\\\ y & =\sqrt{10^{-6}}=10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\end{aligned}$
Substituting value of $y$ in equation (iii),
$ \begin{aligned} \frac{\left[\mathrm{Cu}^{+}\right]}{\left[\mathrm{Ag}^{+}\right]}=\frac{y}{x} & =\frac{10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}}{x}=10^4 \frac{1.0}{1.6} \\ x & =1.6 \times 10^{-7} \end{aligned} $
The concentration of silver ion, $\left[\mathrm{Ag}^{+}\right]=x$ $=1.6 \times 10^{-7} \mathrm{~mol}$ L.
Hence, the value of $x$ in 1.6 $\times 10^{-x}$ is 7 .
KCN, K2SO4, (NH4)2C2O4, NaCl, Zn(NO3)2, FeCl3, K2CO3, NH4NO3 and LiCN
The dissociation constant of a substituted benzoic acid at 25$^\circ$C is 1.0 $\times$ 10$^{-4}$. The pH of a 0.01 M solution of its sodium salt is __________.
Explanation:
Given that
${K_a}({C_6}{H_5}COOH) = 1 \times {10^{ - 4}}$.
pH of 0.01 M ${C_6}{H_5}COONa$.
${K_h} = {{{K_w}} \over {{K_a}}} = {{0.01\,{h^2}} \over {1 - h}}$
$ \Rightarrow {{{{10}^{ - 14}}} \over {{{10}^{ - 4}}}} = {{{{10}^{ - 2}}{h^2}} \over {1 - h}}$
$1 - h$ is approximately equal to 1.
[OH$^-$] = $0.01h$ = 0.01 $\times$ 10$^{-4}$ = 10$^{-6}$
[H$^+$] = 10$^{-8}$
pH = 8
(Note: Degree of dissociation (a) of weak acid and weak base is $<<1;$ degree of hydrolysis of salt $<<1;$ $\left[ {{H^ + }} \right]$ represents the concentration of ${H^ + }$ ions)
| LIST-I | LIST-II | ||
|---|---|---|---|
| P. | (10 mL of 0.1 M NaOH + 20 mL of 0.1 M acetic acid) diluted to 60 mL |
1. | the value of [H+] does not change on dilution |
| Q. | (20 mL of 0.1 M NaOH + 20 mL of 0.1 M acetic acid) diluted to 80 mL |
2. | the value of [H+] changes to half of its initial value on dilution |
| R. | (20 mL of 0.1 M HCL + 20 mL of 0.1 M ammonia solution) diluted to 80 mL |
3. | the value of [H+] changes to two times of its initial value on dilution |
| S. | 10 mL saturated solution of Ni(OH)2 in equilibrium with excess solid Ni(OH)2 is diluted to 20 mL (solid Ni(OH)2 is still present after dilution). |
4. | the value of [H+] changes to ${1 \over {\sqrt 2 }}$ times of its initial value on dilution |
| 5. | the value of [H+] changes to $\sqrt 2 $ times of its initial value on dilution |
Match each process given in LIST-I with one or more effect(s) in LIST-II. The correct option is :
Solubility product constants (K$_{sp}$) of salts of types MX, MX$_2$ and M$_3$X at temperature T are 4.0 $\times$ 10$^{-8}$, 3.2 $\times$ 10$^{-14}$ and 2.7 $\times$ 10$^{-15}$, respectively. Solubilities (mol dm$^{-3}$) of the salts at temperature 'T' are in the order:
2.5 mL of $\frac{2}{5}$M weak monoacidic base (K$_b$ = 1 $\times$ 10$^{-12}$ at 25$^\circ$C) is titrated with $\frac{2}{15}$M HCl in water at 25$^\circ$C. The concentration of H$^+$ at equivalence point is (K$_w$ = 1 $\times$ 10$^{-14}$ at 25$^\circ$C).
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7o C was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is constant (-57.0 kJ/mol), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2) 100 mL of 2.0 M acetic acid (Ka = 2.0 $\times$ 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6o C was measured. (Consider heat capacity of all solutions as 4.2 J/gK and density of all solutions as 1.0 g m/L)
Question
Enthalpy of dissociation (in kJ/mol) of acetic acid obtained from the Expt. 2 is
When 100 mL of 1.0 M KCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7o C was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is constant (-57.0 kJ/mol), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2) 100 mL of 2.0 M acetic acid (Ka = 2.0 $\times$ 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6o C was measured. (Consider heat capacity of all solutions as 4.2 J/gK and density of all solutions as 1.0 g m/L)
Question
The pH of the solution after Expt. 2 is