Question

What is the pH of a solution prepared by dissolving $\mathbf{1}\mathbf{.}\mathbf{23}\mathbf{g}$of

2-nitrophenol (FM 139.11) in 0.250 L?

Short answer

The $\mathrm{PH}$of $1.23g$of 2-nitrophenol in $0.250\mathrm{L}$is$4.34$

Step-by-step solution

Define the formula to find PH.

The formula PH of the solution

$$\begin{gathered} \mathrm{PH}={\mathrm{logA}}_{{\mathrm{H}}^{+}} \\ \mathrm{PH}=\log {\left[{\mathrm{H}}^{+}\right]}_{{\mathrm{yH}}^{+}} \end{gathered}$$

Calculate thePHof 1.23gof 2 -nitrophenol 0.250 L.

Given Formula mass$=139.11$

Acid-equilibria constant,$=139.11$

Mass of 2-nitrophenol$=1.23g$

Volume of 2-nitrophenol$0.250\mathrm{L}$

The formal concentration ${\mathrm{F}}_{\mathrm{HA}}$.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{HA}}=\frac{1.23\mathrm{g}/\left(139.11\mathrm{g}/\mathrm{mol}\right)}{0.250\mathrm{L}} \\ {\mathrm{F}}_{\mathrm{HA}}=0.0354\mathrm{M} \\ \mathrm{HA}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{A}}^{-} \\ \mathrm{F}-\mathrm{x}\frac{{\mathrm{x}}^2}{0.0354-\mathrm{x}}=5.89\times {10}^{-8}\mathrm{M} \\ \mathrm{x}\approx \sqrt{\left(0.0354\right)\left(5.89\times {10}^{-8}\right)} \\ \mathrm{x}=4.57\times {10}^{-5}\mathrm{M} \end{gathered}$$

The of 2-nitrophenol .

$$\begin{gathered} \mathrm{PH}=-\log \left[{\mathrm{H}}^{-}\right] \\ \mathrm{PH}=-\log \left(4.57\times {10}^{-5}\mathrm{M}\right) \\ \mathrm{PH}=4.34 \end{gathered}$$

Thus, the $\mathrm{PH}$of $1.23g$of 2-nitrophenol in $0.250\mathrm{L}$is$4.34$