Question

$n_{CO_{2}} = \frac{224}{22400} = 0.01\text{ and }n_{H^{+}}\text{ used = }\frac{30 \times 1}{1000} = 0.03$

$$\underset{0.01\text{ mole }(L.R.)}{CO_{2} + OH^{-}} \rightarrow \underset{0.01\text{ mole or less}}{HCO_{3}^{-}}$$

Hence, moles of H⁺ used should be 0.01 or less and titration of HCO₃⁻ and H⁺ should not be detected by phenolphthalein. Hence, OH^– must be in excess.

$$\underset{0.01\text{ mole}}{CO_{2}} + \underset{0.02\text{ mole}}{2OH^{-}} \rightarrow \underset{0.01\text{ mole}}{CO_{3}^{2 -}} + H_{2}O$$

Thus, 0.01 mole of CO₃²⁻ will require only 0.01 mole of H+ in the presence of phenolphthalein. As the mole of H⁺ used is 0.03, 0.02 mole OH^– must be present in excess. Hence, total moles of OH^– used = 0.02 + 0.02 = 0.04,

$$\therefore\lbrack NaOH\rbrack_{\text{used}} = \frac{0.04}{1} = 0.04\text{ M}$$