Question

Calculate the volume of 1.01mol/L sodium hydroxide that is required to neutralize 25.00mL of a hydrochloric acid solution that has a pH of 0.32

Answer 1

Answer on Question #73463 – Chemistry – Other

Task:

Calculate the volume of $1.01\mathrm{mol/L}$ sodium hydroxide that is required to neutralize $25.00\mathrm{mL}$ of a hydrochloric acid solution that has a $\mathsf{pH}$ of 0.32

Solution:

$$\begin{array}{l} pH = - \log [H^{+}] = 0.32; \\ [\mathrm{H}^{+}] = 10^{-pH} = 10^{-0.32} = 0.47863\,M; \\ [\mathrm{H}^{+}] = C(HCl) = 0.47863\,M. \end{array}$$

The reaction:

$$NaOH + HCl = NaCl + H_{2}O$$

By the reaction equation:

$$\begin{array}{l} n(NaOH) = n(HCl); \\ C(NaOH) * V(NaOH) = C(HCl) * V(HCl); \\ V(NaOH) = \frac{C(HCl) * V(HCl)}{C(NaOH)} = \frac{0.47863\,M * 25.00\,mL}{1.01\,M} = 11.847; \\ V(NaOH) = 11.85\,mL. \end{array}$$

Answer: $V(NaOH) = 11.85\,mL$

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