Question

what is the maximum concentration of ag+ that can be added to a 0.00780 m solution of na2co3 before a precipitate will form? (ksp for ag2co3 is 8.10 × 10^-12)

Explanation:

Step1: Write the solubility product expression

For \(Ag_2CO_3(s)
ightleftharpoons 2Ag^{+}(aq)+CO_3^{2 - }(aq)\), the solubility - product constant \(K_{sp}=[Ag^{+}]^{2}[CO_3^{2 - }]\)

Step2: Substitute the known values into the \(K_{sp}\) expression

We know that \(K_{sp}=8.10\times 10^{-12}\) and \([CO_3^{2 - }]=0.00780M\) (from \(Na_2CO_3\) which dissociates completely as \(Na_2CO_3 = 2Na^{+}+CO_3^{2 - }\)).

From \(K_{sp}=[Ag^{+}]^{2}[CO_3^{2 - }]\), we can solve for \([Ag^{+}]\). Rearranging the formula gives \([Ag^{+}]=\sqrt{\frac{K_{sp}}{[CO_3^{2 - }]}}\)

Substitute the values: \([Ag^{+}]=\sqrt{\frac{8.10\times 10^{-12}}{0.00780}}\)

First, calculate \(\frac{8.10\times 10^{-12}}{0.00780}=1.03846\times 10^{-9}\)

Then \([Ag^{+}]=\sqrt{1.03846\times 10^{-9}} = 3.22\times 10^{-5}M\)

Answer:

\(3.22\times 10^{-5}M\)