Question

Find the pH if only 6.30 instead of 12.0 mL HCl were added

Short answer

The of Hydrochloric acid was 8.51 if the volume is 6.0ml

Step-by-step solution

Define the Henderson-Hasselbalch Equation.

The relationship between acid pH and pKa in aqueous solutions is described by the Henderson-Hasselbalch equation (acid dissociation constant). This equation can be used to forecast the pH of a buffer solution when the concentrations of such acid as well as its conjugate base, or the base and the related conjugate acid, are determined.

Calculate the pH of Hydrochloric acid.

Consider a basic reaction:

$$\begin{gathered} {\mathrm{BH}}^{+}\stackrel{{\mathrm{K}}_{\mathrm{a}}={\mathrm{K}}_{\mathrm{w}}/{\mathrm{K}}_{\mathrm{b}}}{𝆏}\mathrm{B}+{\mathrm{H}}^{+}, \\ \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left|\mathrm{B}\right|}{\left[{\mathrm{BH}}^{+}\right]} \end{gathered}$$

Where,

${\mathrm{pK}}_{\mathrm{a}}=$acid dissociation constant for weak acid ${\mathrm{BH}}^{+}$

Moles of ${\mathrm{H}}^{+}=\left(0.006\mathrm{L}\right)\left(1.00\mathrm{mol}/\mathrm{L}\right)=0.006\mathrm{molH}+$

$\mathrm{Initial}\mathrm{moles}\begin{array}{cccc}\mathrm{B}& +& {\mathrm{H}}^{+}& \to {\mathrm{BH}}^{+}\\ 0.1206& & 0.006& 0.0296\end{array}$

Final moles 0.0966 - 0.0356

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\mathrm{moles}\mathrm{of}\mathrm{B}}{\mathrm{moles}\mathrm{of}{\mathrm{BH}}^{+}}$

Now

$$\begin{gathered} \mathrm{pH}=8.072+\log \frac{0.0966}{0.0356} \\ \mathrm{pH}=8.51 \end{gathered}$$

Thus, the of Hydrochloric acid was 8.51 if the volume is 6.0ml.