A mixture of 100, m, mol of Ca(OH)₂ and 2, g of sodium sulphate was dissolved in water and volume

English

Gujarati

Hindi

6-2.Equilibrium-II (Ionic Equilibrium)

hard

A mixture of $100\, m\, mol$ of $Ca(OH)_2$ and $2\, g$ of sodium sulphate was dissolved in water and the volume was made upto $100\, mL$. The mass of calcium sulphate formed and the concentration of $OH^-$ in resulting solution, respectively are

(Molar mass of $Ca(OH)_2, Na_2SO_4$ and $CaSO_4$ are $74, 143$ and $136\, g\, mol^{-1}$ respectively; $K_{sp}$ of $Ca(OH)_2$ is $5.5 \times 10^{-6}$)

$1.9\, g, 0.28\, mol\, L^{-1}$

$13.6\, g, 0.28\, mol\, L^{-1}$

$1.9\, g, 0.14\, mol\, L^{-1}$

$13.6\, g, 0.14\, mol\, L^{-1}$

(JEE MAIN-2019)

Solution

$\mathop {\mathop {Ca{{(OH)}_2}}\limits_{100\,m\,mol} }\limits_ – + \mathop {\mathop {N{a_2}S{O_4}}\limits_{14\,m\,mol} }\limits_ – \to \mathop {\mathop {CaS{O_4}}\limits_ – }\limits_{14\,m\,mol} + \mathop {\mathop {2NaOH}\limits_ – }\limits_{28\,m\,mol} $

${w_{CaS{O_4}}} = 14 \times {10^{ – 3}} \times 136 = 1.9\,gm$

$[O{H^ – }] = \frac{{28}}{{100}} = 0.28\,M$

Standard 11

Chemistry

In aqueous solution of $AgCl(s)$, $AgBr(s)$ and $AgI(s)$. Find the ratio $[Cl^- ]$ : $[Br- ]$ : $[I^-]$ at equilibrium if $K_{sp}\, (AgCl)=10^{-10}$, $K_{sp}\, (AgBr) = 10^{-13}$ , $K_{sp}\, (AgI) = 10^{-16}$

hard

Two salts $A _{2} X$ and $MX$ have the same value of solubility product of $4.0 \times 10^{-12}$. The ratio of their molar solubilities i.e. $\frac{S\left(A_{2} X\right)}{S(M X)}=………..$

(Round off to the Nearest Integer).

medium

(JEE MAIN-2021)

On decreasing the $p \mathrm{H}$ from $7$ to $2$ , the solubility of a sparingly soluble salt ($MX$) of a weak acid ($HX$) increased from $10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$ to $10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$. The $p \mathrm{~K}_{\mathrm{s}}$ of $\mathrm{HX}$ is:

normal

(IIT-2023)

There are equal volume of $0.02$ $M$ $CaC{l_2}$ and $0.00004$ $M$ $N{a_2}S{O_4}$ solution are mixed will a precipitation of $CaS{O_4}$ ? ${K_{sp}} = 2.4 \times {10^{ – 5}}$

medium

The solubility in water of a sparingly soluble salt $A{B_2}$ is $1.0 \times {10^{ – 5}}mol\,{l^{ – 1}}$. Its solubility product number will be

medium

(AIEEE-2003)

Solution

Similar Questions

In aqueous solution of $AgCl(s)$, $AgBr(s)$ and $AgI(s)$. Find the ratio $[Cl^- ]$ : $[Br- ]$ : $[I^-]$ at equilibrium if $K_{sp}\, (AgCl)=10^{-10}$, $K_{sp}\, (AgBr) = 10^{-13}$ , $K_{sp}\, (AgI) = 10^{-16}$

Two salts $A _{2} X$ and $MX$ have the same value of solubility product of $4.0 \times 10^{-12}$. The ratio of their molar solubilities i.e. $\frac{S\left(A_{2} X\right)}{S(M X)}=………..$ (Round off to the Nearest Integer).

On decreasing the $p \mathrm{H}$ from $7$ to $2$ , the solubility of a sparingly soluble salt ($MX$) of a weak acid ($HX$) increased from $10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$ to $10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$. The $p \mathrm{~K}_{\mathrm{s}}$ of $\mathrm{HX}$ is:

There are equal volume of $0.02$ $M$ $CaC{l_2}$ and $0.00004$ $M$ $N{a_2}S{O_4}$ solution are mixed will a precipitation of $CaS{O_4}$ ? ${K_{sp}} = 2.4 \times {10^{ – 5}}$

The solubility in water of a sparingly soluble salt $A{B_2}$ is $1.0 \times {10^{ – 5}}mol\,{l^{ – 1}}$. Its solubility product number will be

Start a Free Trial Now

Two salts $A _{2} X$ and $MX$ have the same value of solubility product of $4.0 \times 10^{-12}$. The ratio of their molar solubilities i.e. $\frac{S\left(A_{2} X\right)}{S(M X)}=………..$

(Round off to the Nearest Integer).