JEE MAIN - Chemistry (2023 - 31st January Evening Shift - No. 18)
At $298 \mathrm{~K}$, the solubility of silver chloride in water is $1.434 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}$. The value of $-\log \mathrm{K}_{\mathrm{sp}}$ for silver chloride is _________.
(Given mass of $\mathrm{Ag}$ is $107.9 \mathrm{~g} \mathrm{~mol}^{-1}$ and mass of $\mathrm{Cl}$ is $35.5 \mathrm{~g} \mathrm{~mol}^{-1}$ )
(Given mass of $\mathrm{Ag}$ is $107.9 \mathrm{~g} \mathrm{~mol}^{-1}$ and mass of $\mathrm{Cl}$ is $35.5 \mathrm{~g} \mathrm{~mol}^{-1}$ )
Answer
10
Explanation
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$$ \begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left(\mathrm{Ag}^{+}\right)\left(\mathrm{Cl}^{-}\right)=\mathrm{S} \times \mathrm{S}=\mathrm{S}^2 \\\\ & \mathrm{~S}=\sqrt{\mathrm{K}_{\mathrm{sp}}} \\\\ & \mathrm{S}=\frac{1.434 \times 10^{-3}}{143.4}=10^{-5} \\\\ & \mathrm{~K}_{\mathrm{sp}}=\mathrm{S}^2=10^{-10} \\\\ & \Rightarrow-\log \left(\mathrm{K}_{\mathrm{sp}}\right)=-\log \left(10^{-10}\right)=10 \end{aligned} $$
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