Question

pH of 0.1 M solution of $NaA$ (sodium salt of a weak acid $HA$) is 8.92. Calculate ${pK}_a$ of $HA$. If a drop of HPh $({pK}_{in}=9.52)$ be added to the above solution, predict whether the pink colour will visible or not under the medical fact that our eyes can see the pink colour if the mole % of ionized form of indicator is 25% or more.

• A. Pink colour will be visible
• B. Pink colour will not be visible
• C. Cannot be predicted
• D. None of the above

Answer 1

Answer: (A)

Pink colour will be visible

$\begin{array}{ccccccc}\stackrel{\ominus }{A}& +& H_{2}O& ⟶& HA& +& O\stackrel{\ominus }{H}\\ 0.1& & & & & & \\ 0.1-x& & & & x& & x\end{array}$ $K_r$

$K_r=\frac{x^2}{0.1-x}=\frac{K_w}{K_{HAa}}$

$\Rightarrow K_{aHA}=K_w\left(\frac{0.1-x}{x^2}\right)$

$[p{K}_a=l{K}_w-\log (0.1-x)+2\log x]---\left(1\right)$

$pH=-\log x=8.92\Rightarrow x={10}^{-8.92}$

putting this in $\left(1\right)$ & joining $x$ in $(0.1-x)$

as $x<<0.1$

$p{K}_a=p{K}_w-\log 0.1+2\log x$

$=14-8.92+1$

$=6.08$

$2^{nd}$ part:

$HPh+O\stackrel{\ominus }{H}⟶P\stackrel{\ominus }{H}+H_{2}O$

$P{K}_{in}=9.52\Rightarrow K_{in}=K_{ax}n={10}^{-9.52}$

$K_{rxn}={10}^{-9.52}=\frac{\left[P{h}^{\ominus }\right]\left[H^{\ominus }\right]\left[O{H}^{\ominus }\right]}{\left[HPh\right]\left[O{H}^{\ominus }\right]}$

$\therefore \frac{\left[P{h}^{\ominus }\right]}{\left[HPh\right]}={10}^{-9.52+8.92}$

$={10}^{-0.6}$

$\cong 0.2512>0.25$

$\therefore$ pink variable