Question

We can measure the concentration of HCI solution by reaction with pure sodium carbonate: $\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{+}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\mathbf{\to }\mathbf{2}{\mathbf{Na}}^{\mathbf{+}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{0}\mathbf{+}\mathbf{105}\mathbf{.}\mathbf{9884}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{0007}$required $\mathbf{27}\mathbf{.}\mathbf{35}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{04}\mathbf{mL}$of HCI .

(a) Find the formula mass (and its uncertainty) for ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$.

(b) Find the molarity of the HCI and its absolute uncertainty.

(c) The purity of primary standard ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$is stated to be 99.95 to 100.5wt % , which means that it can react with $\left(100.00\pm 0.05\right)\mathbf{\%}$of the theoretical amount of ${\mathbf{H}}^{\mathbf{+}}$. Recalculate your answer to (b) with this additional uncertainty.

Short answer

⦁ Formula mass and uncertainty of ${\mathrm{Na}}_2{\mathrm{CO}}_3=0.018255\left(\pm 0.093\%\right)\mathrm{mol}$

⦁ Molarity and absolute uncertainty of $\mathrm{HCI}=0.66746\pm 0.001$

⦁ The uncertainty would increase by 5 in the fifth decimal place. Hence, the answer would still be $0.66746\pm 0.001$.

Step-by-step solution

Given

It is given that the concentration of HCI solution can be measured by reaction with pure sodium carbonate

$\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\mathbf{\to }\mathbf{2}{\mathbf{Na}}^{\mathbf{+}}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}$

$$\begin{gathered} \mathbf{m}\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)\mathbf{=}\left(0.9674\pm 0.0009\right)\mathbf{g} \\ \mathbf{M}\mathbf{\left(}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathbf{105}\mathbf{.}\mathbf{9884}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{0007}\mathbf{)}\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{V}\mathbf{\left(}\mathbf{HCI}\mathbf{\right)}\mathbf{=}\left(27.35\pm 0.04\right)\mathbf{mL} \end{gathered}$$

Calculate the number of moles

(a) ${\mathrm{H}}^{+}$ corresponds to HCI

Number of moles of ${\mathrm{Na}}_2{\mathrm{CO}}_3$:

$\mathrm{n}\left({\mathrm{H}}^{+}\right)=2\mathrm{n}\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)$

Calculate the formula mass (and its uncertainty) of  (Na2CO3)

$$\begin{gathered} \mathrm{n}\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)=\frac{\mathrm{m}}{\mathrm{M}} \\ =\frac{0.967\pm 0.0009}{105.9884\pm 0.0007} \\ =\frac{0.967\pm 0.093\%}{105.9884\pm 0.00066\%} \\ =0.0091274\left(\pm 0.093\%\right) \\ \mathrm{n}=2\times \mathrm{n}\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right) \\ =0.018255\left(\pm 0.093\%\right)\mathrm{mol} \end{gathered}$$

Calculate the molarity and absolute uncetainty

(b) The molarity of HCI :

$\mathrm{c}\left(\mathrm{HCI}\right)=\frac{\mathrm{n}}{\mathrm{V}}$

$$\begin{gathered} =\frac{0.018255\left(\pm 0.093\%\right)mol}{0.02735\pm 0.00004L}=0.66740\left(\pm 0.173\%\right) \\ =0.66746\left(\pm 0.173\%\right)=0.66746\left(\pm 0.001155\right)=0.667746\pm 0.001 \end{gathered}$$

Change in uncertainty

(c) In accordance with the question the uncertainty would increase by 5 in the fifth decimal place. Hence, the answer would still be $0.66746\pm 0.001$.