Question

$10.0\mathrm{mL}$ of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ solution is titrated against $0.2\mathrm{M}\mathrm{HCl}$ solution. The following litre values were obtained in $5$ readings $4.8\mathrm{mL},4.9\mathrm{mL},5.0\mathrm{mL},5.0\mathrm{mL}\mathrm{and}5.0\mathrm{mL}$. Based on these readings and the convention of titrimetric estimation the concentration of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ solution is $\mathrm{\_\_\_\_mM}$

Answer 1

Step 1: Given data:

Volume of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ is ${\mathrm{V}}_1=10\mathrm{mL}$

Molarity of ${\mathrm{Na}}_2{\mathrm{CO}}_3$is $M_1=\mathrm{M}$

Since the most precise value from the given readings is $5\mathrm{mL}$; Volume of $\mathrm{HCl}$ is ${\mathrm{V}}_2=5\mathrm{mL}$

Molarity of $\mathrm{HCl}$ is $M_2=0.2\mathrm{M}$

Step 2: Write the balanced equation:

The balanced chemical equation is

${\mathrm{Na}}_2{\mathrm{CO}}_3+2\mathrm{HCl}\to 2\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$

From the above equation, ${\left({\mathrm{n}}_{\mathrm{factor}}\right)}_{N{a}_{2}C{O}_3}=2$

${\left({\mathrm{n}}_{\mathrm{factor}}\right)}_{HCl}=1$

Step 3: Find the molarity of ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$:

According to the Law of Equivalence:

${\mathrm{M}}_{\mathrm{eq}}\mathrm{of}{\mathrm{Na}}_2{\mathrm{CO}}_3={\mathrm{M}}_{\mathrm{eq}}\mathrm{of}\mathrm{HCl}$

$$\begin{gathered} \Rightarrow {\mathrm{M}}_1\times {\mathrm{V}}_1\times {\mathrm{n}}_1={\mathrm{M}}_2\times {\mathrm{V}}_2\times {\mathrm{n}}_2 \\ \Rightarrow \mathrm{M}\times 10\mathrm{mL}\times 2=0.2\mathrm{M}\times 5\mathrm{mL}\times 1 \\ \Rightarrow \mathrm{M}=\frac{0.2\times 5}{2\times 10} \\ \Rightarrow \mathrm{M}=0.05\mathrm{mol}{\mathrm{L}}^{-1} \\ \Rightarrow \mathrm{M}=0.05\times 1000 \\ \Rightarrow \mathrm{M}=50\mathrm{mM} \end{gathered}$$

Hence, the concentration of ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$ solution is $\mathbf{50}\mathbf{}\mathbf{mM}$.