Question

Calculate the concentration of Cd²⁺ resulting from the dissolution of CdCO₃in a solution that is 0.250 Min CH₃CO₂H, 0.375M in NaCH₃CO_2, and 0.010 M in H₂CO₃

Short answer

The concentration of the solution = 3.71.10^-3M.

Step-by-step solution

Define the concentration of the solution

A solution concentration is a measure of the quantity of solute that has been dissolved in a given quantity of solvent or solution. One that contains a relatively high volume of dissolved solute is a concentrated solution

Calculating the concentration of the solution

0.250MCH₃COOH

0.375MCH₃COONa

0.010MH₂CO₃

They calculate the concentration of Cd²⁺resulting from the dissolution ofCdCO_3.

CdCO₃(s) ⇌ Cd²⁺(aq) + CO^2-₃(aq)…

\[{K_{sp}}{\rm{\;for\;CdC}}{{\rm{O}}_3}{\rm{\;is\;}}{K_{sp}}=\left[{{\rm{C}}{{\rm{d}}^{2+}}}\right]\cdot\left[{{\rm{CO}}_3^{2-}}\right]=5.2\cdot{10^{-12}}\]

First, let us take a look at the reaction of ionization of acetic acid

K_afor CH₃COOH is 1.8.10^-5

Dissociate the concentration

The 0.375MCH₃COONa dissociates into 0.375MCH₃COO^- in 0.375MNa⁺, therefore the concentration of H₃O⁺is

\begin{aligned}{{}{}}{{K_a}\left({{\rm{C}}{{\rm{H}}_3}COOH} \right)=\frac{{\left[{{\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^-}}\right]\cdot\left[{{{\rm{H}}_3}{{\rm{O}}^+}}\right]}}{{\left[{{\rm{C}}{{\rm{H}}_3}{\rm{COOH}}}\right]}}}\\{1.8\cdot {{10}^{-5}}=\frac{{0.375\cdot\left[{{{\rm{H}}_3}{{\rm{O}}^+}}\right]}}{{0.250}}}\\{\left[{{{\rm{H}}_3}{{\rm{O}}^+}}\right]=1.2\cdot{{10}^{-5}}{\rm{M}}}\end{aligned}

The carbonization of ion

H₂CO₃(aq) + H₂O(I)⇌ HCO^-₃(aq) + H₃O⁺(aq)

\[{K_{a1}}=\frac{{\left[ {{\rm{HCO}}_3^-}\right]\cdot\left[{{{\rm{H}}_3}{{\rm{O}}^+}}\right]}}{{\left[{{{\rm{H}}_2}{\rm{C}}{{\rm{O}}_3}}\right]}}=4.3\cdot{10^{-7}}\]

HCO₃^-(aq) +H₂O (I)⇌ CO^2-₃ (aq) + H₃O⁺(aq)

\[{K_{a2}}=\frac{{\left[ {{\rm{CO}}_3^{2-}}\right]\cdot\left[{{{\rm{H}}_3}{{\rm{O}}^+}}\right]}}{{\left[{{\rm{HCO}}_3^-}\right]}}=4.7\cdot{10^{ - 11}}\]

 Step 5: The concentration  

The concentration of

\begin{aligned}{{}{}}{{K_{a1}}\cdot{K_{a2}}=\frac{{\left[{{\rm{HCO}}_3^-}\right]\cdot\left[{{{\rm{H}}_3}{{\rm{O}}^+}}\right]}}{{\left[{{{\rm{H}}_2}{\rm{C}}{{\rm{O}}_3}}\right]}}\cdot\frac{{\left[{{\rm{CO}}_3^{2-}}\right]\cdot\left[{{{\rm{H}}_3}{{\rm{O}}^+}}\right]}}{{\left[{{\rm{HCO}}_3^-}\right]}}}\\{{K_{a1}}\cdot{K_{a2}}=\frac{{\left[{{\rm{CO}}_3^{2-}}\right]\cdot{{\left[{{{\rm{H}}_3}{{\rm{O}}^+}}\right]}^2}}}{{\left[{{{\rm{H}}_2}{\rm{C}}{{\rm{O}}_3}}\right]}}}\\{\left[{CO_3^{2-}}\right]=\frac{{{K_{a1}}\cdot{K_{a2}}\cdot\left[{{{\rm{H}}_2}{\rm{C}}{{\rm{O}}_3}}\right]}}{{{{\left[{{{\rm{H}}_3}{{\rm{O}}^+}}\right]}^2}}}}\\{=\frac{{4.3\cdot{{10}^{-7}}\cdot4.7\cdot{{10}^{ -11}}\cdot0.010}}{{{{\left({1.2\cdot{{10}^{-5}}}\right)}^2}}}}\\{=1.40\cdot{{10}^{-9}}{\rm{M}}}\end{aligned}

The concentration

The concentration of

\begin{aligned}{{}{}}{{K_{sp}}=\left[{C{d^{2+}}}\right]\cdot\left[{CO_3^{2-}}\right]}\\{\left[{C{d^{2+}}}\right]=\frac{{{K_{sp}}}}{{\left[{CO_3^{2-}}\right]}}}\\{=\frac{{5.2\cdot{{10}^{-12}}}}{{1.40\cdot{{10}^{-9}}}}}\\{=3.71\cdot{{10}^{-3}}{\rm{M}}}\end{aligned}