Question

An aqueous solution of sodium carbonate,${\mathbf{Na}}_2{\mathbf{CO}}_3$is titrated

with strong acid to a point at which two ${\mathbf{H}}^{+}$ ions have reacted with each

carbonate ion. (a) If 20.0 mL of the carbonate solution reacts with just 40.0 mL

of 0.50 M acid, what is the molarity of the carbonate solution? (b) If the

solution contains 5.0 percent by mass sodium carbonate, what is the density

of the solution? (c) Suppose that you wanted to prepare a liter of an identical

solution by starting with crystalline sodium carbonate decahydrate,${\mathbf{Na}}_2{\mathbf{CO}}_3$?

, rather than with solid${\mathbf{Na}}_2{\mathbf{CO}}_3$ itself. How much of this substance would

you need?

Short answer

$\begin{array}{l}\mathrm{A}\left)\right[{{\mathrm{CO}}_3}^{2-}]=0.5\mathrm{M}\\ \mathrm{B})\mathrm{p}=1.06\mathrm{g}/\mathrm{mL}\\ \mathrm{C})\mathrm{m}=143.24\mathrm{g}\end{array}$

Step-by-step solution

The n of the positive ions of Hydrochloric acid is found

a) If two Hpositiveions react with ${{\mathrm{CO}}_3}^{2-}$, the $n$of${{\text{CO}}_{\text{3}}}^{\text{2-}}$ is:

$\mathrm{n}\left({{\mathrm{CO}}_3}^{2-}\right)=\frac{1}{2}.\mathrm{n}\left(\mathrm{HA}\right)$

The$n$of HA is:

$-\begin{array}{l}\left(\mathrm{n}\right(\mathrm{HA})=[\mathrm{HA}].\mathrm{V}(\mathrm{HA})\\ \mathrm{n}\left(\mathrm{HA}\right)=0.5\mathrm{M}\times 40\times {10}^{(-3)}\\ \mathrm{n}\left(\mathrm{HA}\right)=0.02\mathrm{mol}\end{array}$

The n of Hydrochloric acid is found

The n of is${{\mathrm{CO}}_3}^{2-}$

$\begin{array}{l}\mathrm{n}\left({{\mathrm{CO}}_3}^{2-}\right)=\frac{1}{2}.\mathrm{n}\left(\mathrm{HA}\right)\\ \mathrm{n}\left({{\mathrm{CO}}_3}^{2-}\right)=\frac{1}{2}\times 0.02\mathrm{mol}\\ \mathrm{n}\left({{\mathrm{CO}}_3}^{2-}\right)=0.01\mathrm{mol}\end{array}$

the concentration of${{\mathrm{CO}}_3}^{2-}$

$\begin{array}{l}\mathrm{n}\left({{\mathrm{CO}}_3}^{2-}\right)=\frac{\mathrm{n}\left({{\mathrm{CO}}_3}^{2-}\right)}{\mathrm{Vn}\left({{\mathrm{CO}}_3}^{2-}\right)}\\ \mathrm{n}\left({{\mathrm{CO}}_3}^{2-}\right)=\frac{0.01\mathrm{mol}}{20{.10}^{-3}\mathrm{l}}\\ \mathrm{n}\left({{\mathrm{CO}}_3}^{2-}\right)=0.5\mathrm{M}\end{array}$

The mass of Sodium Carbonate by 5% is found

) If we have$1""\mathrm{L}$of solution, the mass is:

$\begin{array}{l}\mathrm{m}=\mathrm{n}.\left({\mathrm{Na}}_2{\mathrm{CO}}_4\right)\\ \mathrm{m}=0.5\mathrm{mol}\times 105.99\mathrm{g}/\mathrm{mol}\\ \mathrm{m}=53\mathrm{g}\end{array}$

Because the solution contains$5\%$by mass sodium carbonate, the mass is:

$\mathrm{m}=\frac{53\mathrm{g}}{0.05}=1060\mathrm{g}$

The density of the derived solution

The density of solution is

$\begin{array}{l}\mathrm{p}=\frac{\mathrm{m}}{\mathrm{V}}\\ \mathrm{p}=\frac{1060\mathrm{g}}{1000\mathrm{mL}}\\ \mathrm{p}=1.06\mathrm{g}/\mathrm{ml}\end{array}$

The mass of the chemical substance is given

The mass of${\mathrm{Na}}_2{\mathrm{CO}}_3.10{\mathrm{H}}_2\mathrm{O}$ would be

$\begin{array}{l}\mathrm{m}=53\mathrm{g}.\frac{\mathrm{M}({\mathrm{Na}}_2{\mathrm{CO}}_3.10{\mathrm{H}}_2\mathrm{O})}{\mathrm{M}\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)}\\ \mathrm{m}=53\mathrm{g}.\frac{286.2\mathrm{g}/\mathrm{mol}}{105.9\mathrm{g}/\mathrm{mol}}\\ \mathrm{m}=143.24\mathrm{g}\end{array}$

Result

$\begin{array}{l}\mathrm{A}\left)\right[{{\mathrm{CO}}_3}^{2-}]=0.5\mathrm{M}\\ \mathrm{B})\mathrm{p}=1.06\mathrm{g}/\mathrm{mL}\\ \mathrm{C})\mathrm{m}=143.24\mathrm{g}\end{array}$