Solid Ba(NO₃)₂ is gradually dissolved in a 1.0 × 10^{-4}, M, Na₂CO₃ solution. At which conc

English

Gujarati

Hindi

6-2.Equilibrium-II (Ionic Equilibrium)

medium

Solid $Ba(NO_3)_2$ is gradually dissolved in a $1.0 \times 10^{-4}\, M\, Na_2CO_3$ solution. At which concentration of $Ba^{2+}$, precipitate of $BaCO_3$ begins to form ? $(K_{sp}$ for $BaCO_3 = 5.1 \times 10^{-9})$

A

$5.1 \times {10^{ - 5}}\,M$

B

$7.1 \times {10^{ - 8}}\,M$

C

$4.1 \times {10^{ - 5}}\,M$

D

$8.1 \times {10^{ - 7}}\,M$

(JEE MAIN-2013)

Solution

Given $N{a_2}C{O_3} = 1.0 \times {10^{ – 4}}\,M$

$\therefore \,[CO_3^ – ] = 1.0 \times {10^{ – 4}}\,M$

$i.e.\,\,\,s = 1.0 \times {10^{ – 4}}\,M$

At equilibrium

$[B{a^{ + + }}][CO_3^ – ] = {K_{sp}}\,of\,BaC{O_3}$

$[B{a^{ + + }}] = \frac{{{K_{sp}}}}{{[CO_3^ – ]}} = \frac{{5.1 \times {{10}^{ – 9}}}}{{1.0 \times {{10}^{ – 4}}}}$

$ = 5.1 \times {10^{ – 5}}\,M$

Standard 11

Chemistry

Share

Similar Questions

The values of ${K_{sp}}$for $CuS,\,\,A{g_2}S$ and $HgS$ are ${10^{ – 31}},\,{10^{42}}$ and ${10^{ – 54}}$ respectively. The correct order of their solubility in water is

medium

The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively

hard

(JEE MAIN-2020)

The solution is $2$ $L$ of $0.02$ $M$ $NaOH$. The solubility product of $Zn{\left( {OH} \right)_2}$, is $4.5 \times {10^{ – 17}}$. Then how many gram $Zn{\left( {OH} \right)_2}$, maximum soluble in given $NaOH$ solution ?

medium

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 $5.5 \times 10^{-6}$)

hard

(JEE MAIN-2019)

What is the minimum volume of water required to dissolve $1\,g$ of calcium sulphate at $298$ $K$ ? ( For calcium sulphate, $K_{ sp }$ is $9.1 \times 10^{-6}$ ).

medium