Question

Calculate the Ksp of Ag2CrO4. Make sure your Ksp has the correct number of significant digits.

Answer 1

Question #57945, Chemistry / Other

Calculate the $K_{sp}$ of $\mathrm{Ag_2CrO_4}$. Make sure your $K_{sp}$ has the correct number of significant digits.

Solution

According to [1], solubility of argentums chromate at $25^{\circ}\mathrm{C}$ is equal $0.035\mathrm{g/L}$ $\mathrm{H}_2\mathrm{O}$.

Argentums chromate dissociates according to the following equation:

$$\mathrm{Ag_2CrO_4} \leftrightarrow 2\mathrm{Ag^+} + \mathrm{CrO_4^{2-}}$$ $$K_{sp} = [\mathrm{Ag^+}]^2 * [\mathrm{CrO_4^{2-}}]$$

Calculate the concentration of argentums chromate in saturated solution:

$$C(Ag_2CrO_4) = \frac{S(Ag_2CrO_4)}{M(Ag_2CrO_4)} = \frac{0.035\ \mathrm{g/L}}{331.73\ \mathrm{g/mol}} = 1.055 * 10^{-4}\ \mathrm{mol/L}$$

According to the dissociation equation:

$$[\mathrm{Ag^+}] = 2 * C(Ag_2CrO_4) = 2.110 * 10^{-4}\ \mathrm{mol/L}$$ $$[\mathrm{CrO_4^{2-}}] = C(Ag_2CrO_4) = 1.055 * 10^{-4}\ \mathrm{mol/L}$$ $$K_{sp} = (2.110 * 10^{-4})^2 * 1.055 * 10^{-4} = 4.698 * 10^{-12}$$

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