Question

At which value of PH, a 0.010 M mn2+ solution starts to precipitate in the form of Mn(OH)2 and at which value of PH, this Mn2+ solution would precipitate completely([Mn2+]<1.0X10^-5 M)
Ksp[Mn(OH)2]= 2.6X10^-13

Answer 1

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$$\begin{aligned} & \text{Mn}^{2+} + 2OH \rightarrow \text{Mn}(OH)_2 \quad \text{at } 0.03\text{M} \quad \text{Phy}^2 \\ & \text{KSp} = 2.6 \times 10^{-13} \\ & \text{Mn}(OH)_2 \rightarrow \text{Mn}^{2+} + 2OH \\ & \text{KSp} = (S) (2S)^2 \\ & 2.6 \times 10^{-13} = h_5^3 \\ & 2.6 \times 10^{-13} = S^3 \\ & \text{Mn}(OH)_2 \rightarrow \text{Mn}^{2+} + 2OH \\ & (\text{OH})^2 = 5 \times h_5^3 \\ & = 8 \times 10^{-5} \\ & = 5 - \log g \\ & pH = 14 - 4.09 \\ & pH = 14 - 4.03 \\ \end{aligned}$$ $$\begin{aligned} & \text{KSP} = 4.09 \\ & S = 0.65 \times 10^{-13} \\ & S = 6.5 \times 10^{-13} \\ & pH = 4.10^{-5} \\ \end{aligned}$$ $$\begin{aligned} & \text{Mn}^{2+} + 2OH \leftarrow \text{Mn}(OH)_2 \quad \text{at } 0.03\text{M} \quad \text{Phy}^2 \\ & \text{KSp} = 2.6 \times 10^{-13} \\ & \text{Mn}(OH)_2 \rightarrow \text{Mn}^{2+} + 2OH \\ \end{aligned}$$