MarksRiser
MarksRiser
Back to Practice
Chemistry-Solubility Product

NEET Chemistry Solubility Product MCQ Question

Type: MCQ-numerical-Medium-Class 11

For the sparingly soluble salt AgCl, the solubility product constant KspK_{sp} is given as 1.8×10−101.8 \times 10^{-10} at a certain temperature. What is the solubility of AgCl in mol/L in pure water at this temperature?

A

34×10−534 \times 10^{-5}

B

8×10−58 \times 10^{-5}

C

1×10−51 \times 10^{-5}

D

5×10−55 \times 10^{-5}

Correct Answer

Option A

Detailed Explanation

For a sparingly soluble salt like AgCl, the solubility ss is related to its solubility product constant KspK_{sp} by the equation Ksp=s2K_{sp} = s^2. Solving s2=1.8×10−10s^2 = 1.8 \times 10^{-10} gives s=1.8×10−10=1.34×10−5s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5} M.

Found an issue with this question?

Related Questions

Explore More Questions

All Chemistry QuestionsEquilibrium Questions