Question

Find the pH with $\mathbf{1}\mathbf{.}\mathbf{50}{\mathbf{gNaN}}_{\mathbf{2}}\mathbf{P}$instead of 1.20 g.

Short answer

The pH of given${\mathrm{Na}}_2\mathrm{P}$ solution is 5.57.

Step-by-step solution

Concept used.

Henderson-Hassel Balch equation

$$\begin{gathered} \mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{1}}\mathbf{+}\mathbf{log}\frac{\left|{\mathrm{IA}}^{-}\right|}{\left|H_{2}A\right|} \\ \mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{1}}\mathbf{+}\mathbf{log}\frac{\left|A^2\right|}{\left|{\mathrm{HA}}^{-}\right|} \end{gathered}$$

Step2: Calculate the pHwith 1.50NaN2P instead of 1.20 g.

Disodium phthalate${\mathrm{C}}_8{\mathrm{H}}_4{\mathrm{O}}_4{\mathrm{Na}}_2=210\mathrm{g}/\mathrm{mol}$

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_2+\log \frac{\left[{\mathrm{A}}^2\right]}{\left|{\mathrm{HA}}^2\right|}\mathrm{pH} \\ ={\mathrm{pK}}_2+\log \frac{\left[{\mathrm{p}}^2\right]}{\left|\mathrm{HP}-\right|} \\ =5.408+\log \frac{\left(1.50\mathrm{g}\right)/\left(210.094\mathrm{g}/\mathrm{mol}\right)}{\left(1.{00}_{\mathrm{g}}\right)(204.22{\mathrm{g}}_{\mathrm{g}/\mathrm{mol})}} \\ =5.57 \end{gathered}$$

The pH of Na₂P solution as found to be 5.57.