Chemistry - Ionic Equilibrium Question with Solution | TestHub

$[O{H}^{-}]=2.57\times 10^{-11}$ (from previous part, pH = 3.41).

Buffer system: $C{H}_{3}COOH$ / $C{H}_{3}CO{O}^{-}$.

Using the Henderson-Hasselbalch equation:

$$pH=p{K}_a+\log \frac{[C{H}_{3}CO{O}^{-}]}{[C{H}_{3}COOH]}$$

Given $pH=3.41$ and $p{K}_a=4.74$:

$$3.41=4.74+\log \frac{[C{H}_{3}CO{O}^{-}]}{[C{H}_{3}COOH]}$$

Let $x=\frac{[C{H}_{3}CO{O}^{-}]}{[C{H}_{3}COOH]}$.

$$3.41=4.74+\log x$$

$$\log x=3.41-4.74$$

$$\log x=-1.33$$

$$x=10^{-1.33}$$

$$x\approx 0.0468$$

So, $\frac{[C{H}_{3}CO{O}^{-}]}{[C{H}_{3}COOH]}=0.0468$.

For 250 mL of solution, assuming $[C{H}_{3}COOH]=1.00M$:

$[C{H}_{3}CO{O}^{-}]=0.0468M$.

Moles of $C{H}_{3}CO{O}^{-}$ (sodium acetate) needed:

$$moles=0.0468M\times 0.250L=0.0117mol$$

The molar mass of sodium acetate ($C{H}_{3}COONa$) is approximately $82g/mol$.

\Mass of sodium acetate needed:

$$mass=0.0117mol\times 82g/mol\approx 0.96g$$