What volume of KOH would bring the pH to 4.50?

• KOH aqueous (potassium hydroxide solution) is a colourless inorganic liquid that functions as a strong base (alkali).
• KOH solution, commonly known as caustic potash or potash lye, is used in a variety of ways.

It is necessary to calculate the required volume of KOH solution.

Henderson-Hassel Balch equation

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_1+\log \frac{\left[|{\mathrm{HA}}^{‐}\right]}{\left|{\mathrm{H}}_2\mathrm{A}\right|} \\ \mathrm{pH}={\mathrm{pK}}_2+\log \frac{\left[{\mathrm{A}}^2\right]}{\left|{\mathrm{HA}}^{‐}\right]} \end{gathered}$$

Given,

Formula mass =90.035

Above is the ideal pH , the proportion is 1:1 $\left({\mathrm{HOx}}^{‐}{\mathrm{Ox}}^{2‐}\right)$

Are you able to add enough base to convert everything $\left({\mathrm{HOx}}^{‐}{\mathrm{Ox}}^{2‐}\right)$, plus enough extra base to transform the appropriate amount of (HOx- into Ox^2-)

$$\begin{gathered} {\mathrm{H}}_2\mathrm{Ox}+{\mathrm{OH}}^{‐}\to {\mathrm{HOx}}^{-}+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{pH}\approx \frac{1}{2}\left({\mathrm{pK}}_1+{\mathrm{pK}}_2\right)=2.76 \\ {\mathrm{HOx}}^{‐}+{\mathrm{OH}}^{‐}\to {\mathrm{HOx}}^{‐}+{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Determine the value of pH and x value

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_2=4.5 \\ \mathrm{pH}={\mathrm{pK}}_2+\log \frac{\left[{\mathrm{Ox}}^2\right]}{\left[{\mathrm{HOx}}^{\cancel{\mathrm{C}}}|\right]} \\ 4.50=4.166+\log \frac{\mathrm{x}}{0.{0375}_4‐\mathrm{x}} \\ \mathrm{x}=0.{0216}_6\mathrm{mol} \end{gathered}$$

The amount of KOH required is$\left(0.0231\mathrm{mole}\right)/\left(0.800\mathrm{M}\right)=28.7_5\mathrm{mL}$

The total volume KOHneeded to bring the required pHis $47.9+28.7=76.6\mathrm{mL}$

The volume of KOH solution required is discovered to be 76.6 mL