Q19.90P - Chemistry: Molecular Nature Of Matter And Change

Question

When blood is donated, sodium oxalate solution is used to precipitate ${\mathbf{\text{Ca}}}^{\mathbf{\text{2 + }}}$, which triggers clotting. A  $\mathbf{\text{104 mL}}$ sample of blood contains $\mathbf{9}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{g}{\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}\mathbf{/}\mathbf{mL}$. A technologist treats the sample with $\mathbf{\text{100.0 mL}}$  of  ${\mathbf{\text{0.1550 M Na}}}_{\mathbf{\text{2}}}{\mathbf{\text{C}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{4}}}$. Calculate ${\mathbf{\text{[Ca}}}^{\mathbf{\text{2 + }}}\mathbf{\text{]}}$  after the treatment. (See Appendix C for K_sp of ${\mathbf{\text{CaC}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{4}}}\mathbf{·}{\mathbf{\text{H}}}_{\mathbf{\text{2}}}\mathbf{\text{O}}$.)

Step-by-Step Solution

Answer

The amount of ${\text{Ca}}^{\text{2 + }}$after treatment is $\left[{\mathrm{Ca}}^{2+}\right]=3.108\times {10}^{-8}\mathrm{M}$  .

Solubility product

Q_sp- ion-product expression;  Q_sp  value is obtained when the concentrations of ions formed by dissolution of some compound are multiplied. When the solution is saturated Q_sp   value is called K_sp  value (solubility-product constant).

 ${\mathbf{\text{MX}}}_{\mathbf{\text{2}}}\mathbf{\to }{\mathbf{\text{M}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{ + 2X}}}^{\mathbf{\text{ - }}}$

Solid and liquid state is not included in the K_sp equations.

 ${\mathbf{\text{K}}}_{\mathbf{\text{sp}}}{\mathbf{\text{ = [M}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{][X}}}^{\mathbf{\text{ - }}}{\mathbf{\text{]}}}^{\mathbf{\text{2}}}$

S  (Molar solubility) is equal to the concentration of one mol of the ion formed by dissolution of some compound.

Given Data

• Molarity of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ is: $\text{0.155 M}$ 
• Volume of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$  is:  $100\mathrm{mL}=0.1\mathrm{L}$
• Volume of Blood is:  $\text{104 L}$
• Amount of ${\text{Ca}}^{\text{2 + }}$ion in blood is:  $9.7\times {10}^{-5}\mathrm{g}/\mathrm{mL}$

Calculation

First, the dissolution equation for ${\text{CaC}}_{\text{2}}{\text{O}}_4$ –

${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}\text{(s)}\rightleftharpoons {\text{Ca}}^{\text{2 + }}{\text{(aq) + C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2 - }}\text{(aq)}$

Now find the molarity of ${\text{Ca}}^{\text{2 + }}$ions –

$$\begin{gathered} \mathrm{n}\left({\mathrm{Ca}}^{2+}\right)=9.7\times {10}^{-5} \mathrm{g}/\mathrm{ml}\times 104\mathrm{ml}\times \frac{1 \mathrm{mol}}{40.08 \mathrm{g}} \\ \mathrm{n}\left({\mathrm{Ca}}^{2+}\right)=2.5\times {10}^{-4} \mathrm{mol} \\ \mathrm{n}\left({\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right)=0.1\mathrm{L}\times 0.155\mathrm{M} \\ \mathrm{n}\left({\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right)=0.0155 \mathrm{mol} \end{gathered}$$

Now subtract ${\text{n(Ca}}^{\text{2 + }})$ from ${\text{n(C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2 - }})$ –

 $0.0155 \mathrm{mol}-0.00025 \mathrm{mol}=0.01525 \mathrm{mol}$

Now calculate the total volume of solution –

$0.1\mathrm{L}+0.104\mathrm{L}=0.204\mathrm{L}$ 

Now, calculate concentration of  ${\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2 - }}$ ions –

 $$\begin{gathered} \left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]=\frac{0.01525 \mathrm{mol}}{0.204 \mathrm{L}} \\ =7.4\times {10}^{-2}\mathrm{M} \end{gathered}$$

The K_sp value is –

 $$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]\left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right] \\ \left[{\mathrm{Ca}}^{2+}\right]=\mathrm{S} \\ \left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]=7.4\times {10}^{-2} \\ {\mathrm{K}}_{\mathrm{sp}}=2.3\times {10}^{-9} \end{gathered}$$

Number of  ${\text{Ca}}^{\text{2 + }}$ ions left are –

 $$\begin{gathered} 2.3\times {10}^{-9}=\mathrm{S}\times \left(7.4\times {10}^{-2}\right) \\ \left[{\mathrm{Ca}}^{2+}\right]=3.108\times {10}^{-8}\mathrm{M} \end{gathered}$$

Therefore, the value for number of moles is obtained as  $3.108\times {10}^{-8}\mathrm{M}$.