Question

Find the pH of $\mathbf{0}\mathbf{.}\mathbf{002}{\mathbf{MK}}^{\mathbf{+}}{\mathbf{HP}}^{\mathbf{-}}$with Equation$\mathbf{10}\mathbf{-}\mathbf{11}$.

Short answer

The pH of given ${\mathrm{K}}^{+}\mathrm{HP}\text{'}$ solution is 4.28.

Step-by-step solution

Concept used.

Intermediate form of diprotic acid

$\left[H+\right]\mathbf{=}\sqrt{\frac{{\mathbf{K}}_{\mathbf{1}}{\mathbf{K}}_{\mathbf{2}}{\mathbf{I}}^{\mathbf{+}}\mathbf{+}{\mathbf{K}}_{\mathbf{1}}{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{1}}\mathbf{+}\mathbf{F}\mathbf{*}}}$

Step2: Calculate the pH of 0.002MK+HP-.

The pH,

$$\begin{gathered} \left[\mathrm{H}+\right]=\sqrt{\frac{\left(4.7\times {10}^{-3}\right)\left(1.8\times {10}^{-10}\right)\left(0.002\right)+\left(4.7\times {10}^{-3}\right)\left(1.0\times {10}^{-14}\right)}{\left(4.7\times {10}^{-3}\right)+\left(0.002\right)}} \\ \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ =4.28 \end{gathered}$$

The pH of ${\mathrm{K}}^{+}\mathrm{HP}\text{'}$solution is 4.28.