Question

what is the maximum concentration of ni²⁺ that can be added to a 0.00740 m solution of na₂co₃ before a precipitate will form? (k_sp for nico₃ is 1.30 × 10⁻⁷)

Explanation:

Step1: Write the dissolution equation and \(K_{sp}\) expression

The dissolution of \(NiCO_3\) is \(NiCO_3(s)
ightleftharpoons Ni^{2 +}(aq)+CO_3^{2 -}(aq)\). The \(K_{sp}\) expression is \(K_{sp}=[Ni^{2 +}][CO_3^{2 -}]\).

Step2: Determine the concentration of \(CO_3^{2 -}\)

\(Na_2CO_3\) dissociates completely: \(Na_2CO_3 = 2Na^{+}+CO_3^{2 -}\). Given \([Na_2CO_3]=0.00740M\), so \([CO_3^{2 -}]=0.00740M\).

Step3: Solve for \([Ni^{2 +}]\)

Rearrange \(K_{sp}=[Ni^{2 +}][CO_3^{2 -}]\) to \([Ni^{2 +}]=\frac{K_{sp}}{[CO_3^{2 -}]}\). Substitute \(K_{sp}=1.30\times 10^{-7}\) and \([CO_3^{2 -}]=0.00740M\) into the formula: \([Ni^{2 +}]=\frac{1.30\times 10^{-7}}{0.00740}\).

Calculate \(\frac{1.30\times 10^{-7}}{0.00740}\approx1.76\times 10^{-5}M\).

Answer:

\(1.76\times 10^{-5}M\)