The Ksp of AgCl is 1.8 × 10⁻¹⁰ — JEE Chemistry
The $K_{sp}$ of $AgCl$ is $1.8 \times 10^{-10}$. Calculate the molar solubility in pure water and in 0.1 M NaCl solution.
1 Answer
CCamilleDubois28
✓ Accepted
· 2mo ago
▲ 37
In pure water: $s^2 = K_{sp} \Rightarrow s = \sqrt{1.8\times10^{-10}} = 1.34\times10^{-5} M$
In 0.1 M NaCl (common ion):
$$[Ag^+][Cl^-] = K_{sp}$$
$$s \times (0.1 + s) \approx s \times 0.1 = 1.8\times10^{-10}$$
$$s = \frac{1.8\times10^{-10}}{0.1} = 1.8\times10^{-9} M$$
Solubility decreases by a factor of ~$10^4$ — the common ion effect.
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Discussion (3)
RB
Tip for others: you can verify the answer quickly by checking the units.
L
Pro tip: memorise the standard result, it reappears in many problems.
V
How do we know the approximation is valid here?