SATHEE: Equilibrium Question 188

Equilibrium Question 188

Question: The solubility of $A g C l$ in $0.2 M N a C l$ solution $(K_{s p}$ for $A g C l = 1.20 \times 10^{- 10})$ is [MP PET 1996]

Options:

A) $0.2 M$

B) $1.2 \times 10^{- 10}, M$

C) $0.2 \times 10^{- 10}, M$

D) $6 \times 10^{- 10}, M$

Show Answer

Answer:

Correct Answer: D

Solution:

$\underset{a}{A g C l},$ ⇌ $\underset{a}{A g^{+}}, + \underset{a}{C l^{-}},$

$\underset{0.02}{N a C l},$ ⇌ $\underset{0.02,}{N a^{+}}, + \underset{0.02}{C l^{-}},$

$K_{s p} A g C l = 1.20 \times 10^{- 10}$

$K_{s p} A g C l = [A g^{+}] [C l^{-}]$

$= a \times [a + 0.2]$

$= a^{2} + 0.2 a$

$a^{2}$ is a very small so it is a neglected. $K_{s p} A g C l = 0.2 a$

$1.20 \times 10^{- 10} = 0.2 a$

$a = \frac{1.20 \times 10^{- 10}}{0.20} = 6 \times 10^{- 10}$ mole

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Equilibrium Question 188

Question: The solubility of AgCl in 0.2MNaCl solution (Ksp for AgCl=1.20×10−10) is [MP PET 1996]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

Question: The solubility of $A g C l$ in $0.2 M N a C l$ solution $(K_{s p}$ for $A g C l = 1.20 \times 10^{- 10})$ is [MP PET 1996]