Question

A solution is 0.010 M in each of Pb(NO3)2, Mn(NO3)2, and Zn(NO3)2. Solid NaOH is added until the pH of the solution is 8.50. Which of the following statements is true? Salt Ksp Pb(OH)2 1.4 × 10–20 Mn(OH)2 2.0 × 10–13 Zn(OH)2 2.1 × 10–16 a. All three hydroxides will precipitate. b. No precipitate will form. c. Only Pb(OH)2 will precipitate. d. Only Mn(OH)2 will precipitate. e. Only Zn(OH)2 and Pb(OH)2 will precipitate.

Answer 1

Answer:

All three hydroxides will precipitate

Explanation:

General solubility equilibrium of bivalent metal hydroxides is represented as-

$M(OH)_{2}\rightleftharpoons M^{2+}+2OH^{-}$

solubility product, $K_{sp}=[M^{2+}][OH^{-}]^{2}$

If , at pH 8.50, product of $[M^{2+}]$ and $[OH^{-}]^{2}$ exceeds $K_{sp}$ values of given hydroxides then hydroxides of given metal ions will precipitate.

pH = 8.50

or, pOH = 14-8.50 = 5.50

or, $-log[OH^{-}]$ = 5.50

or, $[OH^{-}]$ = $10^{-5.50}$

or, $[OH^{-}]$ = $3.16\times 10^{-6}$

As each salt solutions have concentration of 0.010 M therefore concentrations of each metal ions ($M^{2+}$) are 0.010 M

So, $[M^{2+}][OH^{-}]^{2}=0.010\times (3.16\times 10^{-6})^{2}=3.16\times 10^{-8}$

As product of $[M^{2+}]$ and $[OH^{-}]^{2}$ exceeds $K_{sp}$ values of every hydroxides therefore all three hydroxides will precipitate.