Question

10 gm of impure sodium carbonate is dissolved in water and the solution is made up to 250 ml. To 50 ml of this made up solution, 50 ml of 0.1N, HCl is added and the mixture after shaking well required 10 ml of 0.16N, NaOH solution for complete titration. Calculate the % purity of the sample.

Answer 1

$N{a}_{2}C{O}_3+HCl\to NaCl+H_{2}O+C{O}_2$

$HCl+NaOH\to NaCl+H_{2}O$

Equivalents of HCl reacted with $NaOH=1.6\times {10}^{–3}$

So, for 50 ml solution, equivalents of HCl reacted with $N{a}_{2}C{O}_3=5\times {10}^3–1.6\times {10}^{–3}=3.4\times {10}^{–3}$

So, for 250 ml solution, equivalents of HCl reacted with $N{a}_{2}C{O}_3=5\times 3.4\times {10}^{–3}$

So, equivalents of $N{a}_{2}C{O}_3=5\times 3.4\times {10}^{–3}$

$\frac{\text{mass of}N{a}_{2}C{O}_3}{\frac{106}{2}}=5\times 3.4\times {10}^{-3}$

So $\%=\frac{0.901}{10}\times 100=9\%$