The solubility product of a MA is 2×10–12 at 25∘C. A solution is 0.1 M in M(NO3)2 and it is saturated with 0.01 M H2A then the minimum pH that must be maintained to start precipitation of MA.Given: Ka1(H2A)=4×10–7 at 25∘CKa2(H2A)=5×10–11 at 25∘C

The solubility product of a MA is 2×10–12 at 25∘C. A solution is

1.	The solubility product of a MA is 2×10–12 at 25∘C. A solution is 0.1 M in M(NO3)2 and it is saturated with 0.01 M H2A then the minimum pH that must be maintained to start precipitation of MA.Given: Ka1(H2A)=4×10–7 at 25∘CKa2(H2A)=5×10–11 at 25∘C
Answer» The solubility product of a MA is $2 \times 10^{- 12}$ at $25^{\circ} C$ . A solution is $0.1 M$ in $M (N O_{3})_{2}$ and it is saturated with $0.01 M H_{2} A$ then the minimum pH that must be maintained to start precipitation of $M A$ . Given: $K_{a_{1}} (H_{2} A) = 4 \times 10 - 7$ at $25^{\circ} C$ $K_{a_{2}} (H_{2} A) = 5 \times 10 - 11$ at $25^{\circ} C$

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The solubility product of a MA is 2×10–12 at 25∘C. A solution is 0.1 M in M(NO3)2 and it is saturated with 0.01 M H2A then the minimum pH that must be maintained to start precipitation of MA.Given: Ka1(H2A)=4×10–7 at 25∘CKa2(H2A)=5×10–11 at 25∘C​​​​​​​

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The solubility product of a MA is 2×10–12 at 25∘C. A solution is 0.1 M in M(NO3)2 and it is saturated with 0.01 M H2A then the minimum pH that must be maintained to start precipitation of MA.Given: Ka1(H2A)=4×10–7 at 25∘CKa2(H2A)=5×10–11 at 25∘C