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Ionic Equilibrium question

2023 · 13 Apr · Shift 1 · Q22

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JEE Main
Chemistry
Ionic Equilibrium
2023 · 13 Apr · Shift 1 · Q22

Ionic Equilibrium question

2023 · 13 Apr · Shift 1 · Q22

JEE MainChemistryIonic EquilibriumNumerical+4 / −1

25.0 \mathrm{~mL}

of

0.050 ~\mathrm{M} ~\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}

is mixed with

25.0 \mathrm{~mL}

of

0.020 ~\mathrm{M} ~\mathrm{NaF} . \mathrm{K}_{\mathrm{Sp}}

of

\mathrm{BaF}_{2}

is

0.5 \times 10^{-6}

at

298 \mathrm{~K}

. The ratio of

\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{F}^{-}\right]^{2}

and

\mathrm{K}_{\mathrm{sp}}

is

\underline{\hspace{2cm}}

. (Nearest integer)

Numerical answer

View written solutionFree

Correct answer: 5

Write the reaction and identify ions after mixing

We mix:

$25.0\,\text{mL}$ of $0.050\,\text{M }\text{Ba(NO}_3)_2$
$25.0\,\text{mL}$ of $0.020\,\text{M }\text{NaF}$

Total volume after mixing: $V_{\text{total}}=25.0+25.0=50.0\,\text{mL}=0.0500\,\text{L}$

Dissociation: $\text{Ba(NO}_3)_2 \rightarrow \text{Ba}^{2+}+2\text{NO}_3^-$ $\text{NaF} \rightarrow \text{Na}^+ + \text{F}^-$

Possible precipitate: $\text{Ba}^{2+}+2\text{F}^- \rightleftharpoons \text{BaF}_2(s)$

Calculate initial moles of ions

Moles of $\text{Ba}^{2+}$ : $n_{\text{Ba}^{2+}}=0.050\times 0.0250=1.25\times 10^{-3}\,\text{mol}$

Moles of $\text{F}^-$ : $n_{\text{F}^-}=0.020\times 0.0250=5.0\times 10^{-4}\,\text{mol}$

Find concentrations just after mixing (before any precipitation)

$[\text{Ba}^{2+}] = \frac{1.25\times 10^{-3}}{0.0500}=0.025\,\text{M}$

$[\text{F}^-] = \frac{5.0\times 10^{-4}}{0.0500}=0.010\,\text{M}$

Calculate ionic product

For $\text{BaF}_2$ , $Q=[\text{Ba}^{2+}][\text{F}^-]^2$

So, $Q=(0.025)(0.010)^2$ $Q=0.025\times 10^{-4}=2.5\times 10^{-6}$

Compare with solubility product

Given: $K_{sp}=0.5\times 10^{-6}=5.0\times 10^{-7}$

Required ratio: $\frac{[\text{Ba}^{2+}][\text{F}^-]^2}{K_{sp}}=\frac{2.5\times 10^{-6}}{0.5\times 10^{-6}}=5$

Final answer

Nearest integer: $\boxed{5}$

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