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aivan3 [116]
4 years ago
12

A solution contains 1.32×10-2 M lead nitrate and 1.16×10-2 M copper(II) acetate. Solid sodium carbonate is added slowly to this

mixture.
A. What is the formula of the substance that precipitates first? formula =
B. What is the concentration of carbonate ion when this precipitation first begins? [CO32-] = M
Chemistry
1 answer:
user100 [1]4 years ago
8 0

Answer:

CuCO₃ is the formula of the substance that will precipitate out first

the concentration of carbonate ion when this precipitation first begins = 1.212 × 10⁻¹¹ M

Explanation:

Pb(NO_3)_2 \ + \ Na_2CO_3 -----> PbCO_3 \ + \ 2NaNO\\\\ksp \ of \ PbCO_3 =  1.6*10^{-13} \\\\ \\Cu (CH_3COO)_2 \ + \ Na_2CO_3 ------> CuCO_3 \ + \ Na(C_2H_3O_2)\\\\ksp \ of  CuCO_3 = 1.3*10^{-10}\\\\As \ ksp \ of CuCO_3 > PbCO_3\\\\Then \ \ CuCO_3 \ will \ precipitate \ out \ first

PbCO_3 ---->   Pb^{2+}_{(aq)} + CO^{2-}_{3(aq)}

[Pb^{2+}] will be [Pb(NO_3)_2] = 1.32 *10^{-2} \ M

ksp =  [Pb^{2+}][CO^{2-}_3]

1.6*10^{-13} = (1.32*10^{-2})(CO^{2-}_3)\\\\(CO^{2-}_3) = \frac{1.6*10^{-13}}{(1.32*10^{-2})}\\\\(CO^{2-}_3) = 1.212 *10^{-11} \ \ M

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