If solubility product of BaS{O₄} is 1.5 × {10^{ - 9}} in water, its solubility in moles per litre

English

Gujarati

Hindi

6-2.Equilibrium-II (Ionic Equilibrium)

medium

If the solubility product of $BaS{O_4}$ is $1.5 \times {10^{ - 9}} $ in water, its solubility in moles per litre, is

A

$1.5 \times {10^{ - 9}}$

B

$3.9 \times {10^{ - 5}}$

C

$7.5 \times {10^{ - 5}}$

D

$1.5 \times {10^{ - 5}}$

Solution

(b) $BaS{O_4}$ $ \rightleftharpoons $ $B{a^{2 + }} + SO_4^{ – \, – }$

Solubility constant $ = S \times S$

$1.5 \times {10^{ – 19}} = {S^2}$;$S = \sqrt {1.5 \times {{10}^{ – 19}}} $;$S = 3.9 \times {10^{ – 5}}$

Standard 11

Chemistry

Share

Similar Questions

The solubility products of $Al(OH)_3$ and $Zn(OH)_2$ are $8.5 \times 10^{-23}$ and $1.8 \times 10^{-14}$ at room temperature. If the solution contains $Al^{3+}$ and $Zn^{2+}$ ions, the ion first precipitated by adding $NH_4OH$ is

hard

Any precipitate is formed when

medium

(AIIMS-1982)

${K_{sp}}$ of $Pb$ $I_{2}$ is $ 1.4 \times {10^{ – 8}}$. The molecular mass of $Pb$ $I_{2}$ is $461$ $g$ $mol^{-1}$ Then molecular mass of $Pb{\left( {N{O_3}} \right)_2}$ is $331.9$ $mol^{-1}$ So, $(a)$ In $500$ $mL$ water $(b)$ $500$ $mL$ $0.10$ $M$ $KI$ $(c)$ What is the weight of $PbI_{2}$ when soluble in $1.33$ $g$ $Pb{\left( {N{O_3}} \right)_2}$ containing $500$ $mL$ solution ?

hard

The compound insoluble in acetic acid is

easy

(IIT-1986)

Solubility of $BaF_2$ in a solution of $Ba(NO_3)_2$ will be represented by the concentration term?

medium