Question

Solid NaBr is slowly added to a solution that is 0.010 M inCu+ and 0.010 M in Ag+. (a) Which compoundwill begin to precipitate first? (b) Calculate [Ag+] when CuBr justbegins to precipitate. (c) What percent of Ag+ remains in solutionat this point?
a) AgBr: \(\displaystyle{\left({0.010}+{s}\right)}{s}={4.2}\cdot{10}^{{-{8}}}\) \(\displaystyle{s}={4.2}\cdot{10}^{{-{9}}}{M}{B}{r}\) needed form PPT
CuBr: \(\displaystyle{\left({0.010}+{s}\right)}{s}={7.7}\cdot{\left({0.010}+{s}\right)}{s}={7.7}\cdot{10}^{{-{13}}}\) Ag+=\(\displaystyle{1.8}\cdot{10}^{{-{7}}}\)
b) \(\displaystyle{4.2}\cdot{10}^{{-{6}}}{\left[{A}{g}+\right]}={7.7}\cdot{10}^{{-{13}}}\) [Ag+]\(\displaystyle={1.8}\cdot{10}^{{-{7}}}\)
c) \(\displaystyle{\frac{{{1.8}\cdot{10}^{{-{7}}}}}{{{0.010}{M}}}}\cdot{100}\%={0.18}\%\)

Answer 1

Yes ur correct.i a and b part
c ) 0.0018% not 0.18%