The K s p of A g 2 C r O 4 is 1.1 × 1 0 - 12 at 0.1 K . The solubility (in m o l / L ) of A g 2 C r O 4 in a 0.1 M A g N O 3 solution is

Ag 2 CrO 4 s ⇌ 2 Ag + aq 2 s + 0.1 + CrO 4 2 − aq s K s p = ( 0.1 ) 2 s = 1.1 × 1 0 - 12 s = 1.1 × 1 0 - 10 M

Start Free Mock LoginAlready registered? Login here

Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductEasy2 minPYQ_2020

ChemistryEasysingle choice

The $K_{s p}$ of $A g_{2} C r O_{4}$ is $1.1 \times 1 0^{- 12}$ at $0.1 K$ . The solubility (in $m o l / L$ ) of $A g_{2} C r O_{4}$ in a $0.1 M A g N O_{3}$ solution is

Options:

Answer:

Solution:

${Ag}_{2} {CrO}_{4} (s) ⇌ \underset{2 s + 0.1}{2 {Ag}^{+} (aq)} + \underset{s}{{CrO}_{4}^{2 -} (aq)}$
$K_{s p} = (0.1)^{2} s = 1.1 \times 1 0^{- 12}$
$s = 1.1 \times 1 0^{- 10} M$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...