Question

Calculate the volume of 1.50MCH₃CO₂H required to dissolve a precipitate composed of 350mg each of and CaCO_3,SrCO_3,BaCO₃

Short answer

The volume of the dissolve precipitate is 10.19mL.

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 volume of the dissolved precipitate

calculate the volume of 1.50MCH₃CO₂H required to dissolve a precipitate composed of

350mg of CaCO₃(0.35g)

350mg ofSrCO₃(0.35g)

350mg ofBaCO₃ (0.35g)

\begin{aligned}{{n_{{\rm{CaC}}{{\rm{O}}_3}}}=\frac{{0.35{\rm{g}}}}{{{M_{CaC{O_3}}}}}=\frac{{0.35{\rm{g}}}}{{100.0869{\rm{g}}/{\rm{mol}}}}=3.50\cdot{{10}^{-3}}{\rm{mol}}}\\{{n_{SrC{O_3}}}=\frac{{0.35g}}{{{M_{SrC{O_3}}}}}=\frac{{0.35{\rm{g}}}}{{147.63{\rm{g}}/{\rm{mol}}}}=2.37\cdot{{10}^{-3}}{\rm{mol}}}\\{{n_{{\rm{BaC}}{{\rm{O}}_3}}}=\frac{{0.35{\rm{g}}}}{{{M_{BaC{\rm{C}}{{\rm{O}}_3}}}}}=\frac{{0.35{\rm{g}}}}{{197.34{\rm{g}}/{\rm{mol}}}}=1.77\cdot{{10}^{-3}}{\rm{mol}}}\end{aligned}

The reaction of the solution

The total number of moles of ion, that will react is

\[{n_{{\rm{C}}{{\rm{O}}_3}^{2 - }}}={n_{{\rm{CaC}}{{\rm{O}}_3}}}+{n_{{\rm{SrC}}{{\rm{O}}_3}}}+{n_{{\rm{BaC}}{{\rm{O}}_3}}}=7.64\cdot{10^{-3}}{\rm{mol}}\]

The reaction betweenCO₃2 – CH₃CO₂H.

CO₃^2- (aq) + CH₃CO₂H(aq) ⇌ HCO₃^- (aq) + CH₃CO₂^-

\[{K_c}=\frac{{\left[{HC{O_3}-}\right]\cdot\left[{C{H_3}CO_2^-}\right]}}{{\left[{C{O_3}^{2-}}\right]\cdot\left[{C{H_3}C{O_2}H}\right]}}\]

\[{K_c}=\frac{{\left[{HC{O_3}-}\right]\cdot\left[{C{H_3}CO_2^-}\right]}}{{\left[{C{O_3}^{2-}}\right]\cdot\left[{C{H_3}C{O_2}H}\right]}}\]

 Acetic acid dissolution

let us take a look at acetic acid dissociation

CH₃CO₂H (aq) + H₂O ⇌ H₃O⁺ (aq) + CH₃ CO₂^- (aq)

\[{{\rm{K}}_a}=\frac{{\left[ {{{\rm{H}}_3}{{\rm{O}}^+}}\right]\cdot\left[{{\rm{C}}{{\rm{H}}_3}{\rm{CO}}_2^-}\right]}}{{\left[{{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}}\right]}}=1.8\cdot{10^{-5}}\]

\[\left[ {{\rm{C}}{{\rm{H}}_3}{\rm{CO}}_2^-}\right]=\frac{{1.8\cdot{{10}^{-5}}\cdot\left[{{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}}\right]}}{{\left[{{{\rm{H}}_3}{{\rm{O}}^+}}\right]}}\]

let us take a look at dissociation of HCO₃^-

HCO₃^-(aq) + H₂O (I) ⇌ H₃O⁺(aq) + CO₃²⁺ (aq)

\[{K_a}=\frac{{\left[{{H_3}{O^+}}\right]\cdot\left[{CO_3^{2-}}\right]}}{{\left[{HCO_3^-}\right]}}=4.7\cdot{10^{-11}}\]

\[\left[{CO_3^{2-}}\right]=\frac{{4.7\cdot{{10}^{-11}}\cdot\left[{HCO_3^-}\right]}}{{\left[{{H_3}{O^+}}\right]}}\]

The substitute reaction

Now we can substitute these results in equation

\begin{aligned}{{K_c}=\frac{{\left[{{\rm{HCO}}_3^-}\right]\cdot\left[{{\rm{C}}{{\rm{H}}_3}{\rm{CO}}_2^-}\right]}}{{\left[{{\rm{C}}{{\rm{O}}_3}^{2-}}\right]\cdot\left[{{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}}\right]}}}\\{=\frac{{\left[{{\rm{HCO}}_3^-}\right]\cdot\frac{{\left.{1.8\cdot{{10}^{-5}}\cdot{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}}\right]}}{{\left[{{{\rm{H}}_3}{{\rm{O}}^+}}\right]}}}}{{\frac{{4.7\cdot{{10}^{-11}}\cdot\left[{{\rm{HCO}}_3^-}\right]}}{{\left[ {{{\rm{H}}_3}{{\rm{O}}^+}}\right]}}\cdot\left[{{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}}\right]}}}\\{=\frac{{1.8\cdot{{10}^{-5}}}}{{4.7\cdot{{10}^{-11}}}}}\\{=3.83\cdot{{10}^5}}\end{aligned}

SinceK_c is too large, the concentration of the product is 3.38 .10⁵ times larger than the concentration of reactant. In other words, reaction will go to completion

Hence 7.64. 10^-3 moles of CO₃^2- will react with 7.64. 10^-3 moles of CH₃CO₂Hand produce7.64. 10^-3 moles of HCO₃ -

Reaction to the number of moles

We have to find the number of moles of CH₃CO₂Hthat will react at 7.64. 10^-3 moles of HCO₃ -

HCO₃^-(aq) + CH₃CO₂H(aq)⇌ H2CO₃(aq) + CH₃CO₂^-

\[{K_c}=\frac{{\left[ {{H_2}C{O_3}}\right]\cdot \left[ {C{H_3}CO_2^-}\right]}}{{\left[{HCO_3^-}\right]\cdot\left[{{\rm{C}}{{\rm{H}}_3}C{O_2}H}\right]}}\]

The dissociation of H₂CO₃

HCO₃^-(aq) + H₂O (I)⇌H₃O⁺(aq) + HCO₃²⁺ (aq)

\[{K_a}=\frac{{\left[{{H_3}{O^+}}\right]\cdot\left[{HCO_3^-}\right]}}{{\left[{{H_2}C{O_3}}\right]}}=4.3\cdot{10^{-7}}\]

\[\left[{HCO_3^-}\right]=\frac{{4.3\cdot{{10}^{-7}}\cdot\left[{{H_2}C{O_3}}\right]}}{{\left[ {{H_3}{O^+}}\right]}}\]

By substituting the  H2CO3

\[\begin{aligned}{{K_c}=\frac{{\left[{{{\rm{H}}_2}{\rm{C}}{{\rm{O}}_3}}\right]\cdot\left[{{\rm{C}}{{\rm{H}}_3}{\rm{CO}}_2^-}\right]}}{{\left[{{\rm{HCO}}_3^-}\right]\cdot\left[{{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}}\right]}}}\\{=\frac{{\left[{{{\rm{H}}_2}{\rm{C}}{{\rm{O}}_3}}\right]\cdot\frac{{\left.{1.8\cdot{{10}^{-5}}\cdot{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}} \right]}}{{\left[{{{\rm{H}}_3}{{\rm{O}}^+}}\right]}}}}{{\frac{{4.3\cdot{{10}^{-7}}\cdot\left[{{{\rm{H}}_2}{\rm{C}}{{\rm{O}}_3}}\right]}}{{\left[{{{\rm{H}}_3}{{\rm{O}}^+}}\right]}}\cdot\left[{{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}}\right]}}}\\{=\frac{{1.8\cdot{{10}^{-5}}}}{{4.3\cdot{{10}^{-7}}}}}\\{=41.86}\end{aligned}

Finding the total amount of the moles

Since \[{K_c}\] is too large, the concentration of the product is \[41.86\] times larger than the concentration of the reactant. In other words, the reaction will go to completion

Hence the \[7.64 \cdot {10^{-3}}\]moles of \[{\rm{HC}}{{\rm{O}}_3}\] will react with\[7.64 \cdot {10^{-3}}\] moles of \[{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}\]and produce\[7.64 \cdot {10^{-3}}\]\[{{\rm{H}}_2}{\rm{C}}{{\rm{O}}_3}\]

The total number of moles\[{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}\]\[15.28 \cdot {10^{-3}}{\rm{\;moles\;}}\left( {7.64\cdot {{10}^{-3}}}\right.\]reacted with\[{\rm{C}}{{\rm{O}}_3}^{2-}{\rm{\;and\;}}7.64\cdot {10^{-3}}\]reacted with\[\left. {{\rm{HC}}{{\rm{O}}_3}^-} \right)\]

The volume required \[1.50{\rm{MC}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}\]

\begin{aligned}{\left[{{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2}{\rm{H}}}\right]=\frac{n}{V}}\\{V=\frac{n}{{\left[{C{H_3}C{O_2}H}\right]}}}\\{=\frac{{15.28\cdot{{10}^{-3}}{\rm{mol}}}}{{1.50{\rm{mol}}/{\rm{L}}}}}\\{=10.19\cdot{{10}^{-3}}{\rm{L}}}\\{=10.19{\rm{mL}}}\end{aligned}