Question

A 25.00 - mL solution containing 0.031 10 M ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}$was titrated with 0.025 70 M $\mathbf{Ca}\mathbf{(}{\mathbf{NO}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}$to precipitate calcium oxalate:$\mathit{C}{\mathit{a}}^{\mathbf{2}\mathbf{+}}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\to }\mathit{C}\mathit{a}\mathit{C}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$ Find pCa²⁺ at the following volumes

of :$\mathit{C}\mathit{a}\mathbf{(}{\mathbf{NO}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}$

(a) 10.00,

(b) Ve

(c) 35.00 mL

Short answer

1. The ${\mathrm{pCa}}^{2+}$value for the titration of ${\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_2\mathrm{vs}\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$is 6.81
2. The ${\mathrm{pCa}}^{2+}$value at equivalence point for the titration of ${\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_2\mathrm{vs}\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$is 4.32

3. The ${\mathrm{pCa}}^{2+}$value after equivalence point for the titration of ${\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_2\mathrm{vs}\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$is 2.69

Step-by-step solution

Define The molarity and formula to be used.

The ratio of moles of solute (in gram) to the volume of the solution is known as molarity (in litres). The formula for calculating a solution's molarity is:

Molarity $\left(M\right)=\frac{\mathrm{Moles}\mathrm{of}\mathrm{solute}\left(\mathrm{ing}\right)}{\mathrm{Volume}\mathrm{of}\mathrm{solution}(\mathrm{in}L)}$

Step 2: Calculate the titration of  Na2C2O2 vs  Ca(NO3)2

Given,

25.00mL of $0.03110{\mathrm{MNa}}_2{\mathrm{C}}_2{\mathrm{O}}_2$

10.00mL of $0.02570\mathrm{MCa}{\left({\mathrm{NO}}_3\right)}_2$

The titration reaction of ${\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_2\mathrm{vs}\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$

localid="1663396150098" ${\mathrm{Ca}}^{2+}+{\mathrm{C}}_2{\mathrm{O}}_4^{-}\to {\mathrm{CaCO}}_{4\left(s\right)}$

There are more moles of oxalate ion in solution than calcium ions before the equivalency threshold.

The unprecipitated oxalate ion's strength is estimated as

Moles of $C_2{\mathrm{O}}_4^{-}$original moles of $C_2{\mathrm{O}}_4^{-}$- moles of $C{a}^{2+}$added

$$\begin{gathered} =\left(0.025\mathrm{L}\right)\left(0.03110\mathrm{mol}/\mathrm{L}\right)-\left(0.010\mathrm{L}\right)\left(0.02570\mathrm{mol}/\mathrm{L}\right) \\ =0.00005205\mathrm{mol}{\mathrm{C}}_2{\mathrm{O}}_4^{-} \end{gathered}$$

Total volume of both solution is 0.035L(25.00mL + 10.00m)

The concentration of iodide ion is

$$\begin{gathered} \left[{\mathrm{C}}_2{\mathrm{O}}_4^{-}\right]=\frac{0.0005205\mathrm{mol}}{0.035\mathrm{L}} \\ =0.01487\mathrm{M} \end{gathered}$$

The silver ion concentration that is in equilibrium with the iodide ion is

$$\begin{gathered} \left[C{a}^{2+}\right]=\frac{K_{sp}}{\left|C_2{O}_4\right|} \\ =\frac{2.3\times {10}^{17}}{0.01487} \\ =1.5466\times {10}^{-7}pC{a}^{2+} \\ =-{\log }_{10}\left[C{a}^{2+}\right] \\ =-\log \left(1.5466\times {10}^{-7}\right) \\ =6.81 \end{gathered}$$

Step 3: Calculate the equivalence point for the titration.

Given,

25.00 mL of 0.03110 M ${\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_4$

$0.02570\mathrm{MCa}{\left({\mathrm{NO}}_3\right)}_2$

The titration reaction of ${\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_2\mathrm{vs}\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$

role="math" localid="1663393750871" ${\mathrm{Ca}}^{2+}+{\mathrm{C}}_2{\mathrm{O}}_4^{-}\to {\mathrm{CaCO}}_4\left(s\right)$

The moles of oxalate in solution equal the calcium ions at the equivalency point. The silver ion and iodide ion concentrations are computed as role="math" localid="1663395559816" $\left[{\mathrm{Ca}}^{2+}\right]=\left[{\mathrm{C}}_2{\mathrm{O}}_4^{-}\right]=K_{sp}$

$$\begin{gathered} \left(x\right)\left(x\right)=2.3\times {10}^{-9} \\ x=\sqrt{2.3\times {10}^{-1}} \\ =4.795\times {10}^{-5}pC{a}^{2+} \\ =-{\log }_{10}\left[C{a}^{2+}\right] \\ =-\log \left(4.795\times {10}^{-5}\right) \\ =4.32 \end{gathered}$$

Step 4: Calculate the after-equivalence point for the titration.

Given,

25.00mL of 0.03110 M${\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_4$

$0.02570\mathrm{MCa}{\left({\mathrm{NO}}_3\right)}_2$

The titration reaction of ${\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_2\mathrm{vs}\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$

${\mathrm{Ca}}^{2+}+{\mathrm{C}}_2{\mathrm{O}}_4^{-}\to {\mathrm{CaCO}}_4\left(s\right)$

The volume of calcium ions at equivalence point is

$$\begin{gathered} \left(0.025\mathrm{L}\right)\left(0.03110\mathrm{M}\right)={\mathrm{V}}_{\mathrm{e}}\times \left(0.02570\right) \\ \left(0.025\mathrm{L}\right)\left(0.03110\mathrm{M}\right)={\mathrm{V}}_{\mathrm{e}}\times \left(0.02570\right) \end{gathered}$$

Most of oxalate ion are precipitated at equivalence point. After equivalence point, more moles of calcium ions present in solution. The volume of calcium ion after equivalence point is$35.00-30.25=4.75\mathrm{mL}$

Moles of${\mathrm{Ca}}^{2+}$

$\begin{array}{l}=(0.004751\mathrm{L})\left(0.02570\mathrm{mol}/\mathrm{L}\right)\\ =0.000344\mathrm{mol}{\mathrm{Ca}}^{2+}\end{array}$

$$\begin{gathered} \left[C{a}^{2+}\right]=\left(0.0001221mol\right)/\left(0.060L\right) \\ =0.0020346 \\ pC{a}^{2+}=-{\log }_{10}\left[C{a}^{2+}\right] \\ =-\log \left(0.0020346\right) \\ =2.69 \end{gathered}$$

Thus, the after-value equivalence point is 2.69