Question

LE

Answer 1

Answer:

pH=2.16

Explanation:

Hello,

In this case, the first step is to compute the molarity of the 0.10%-w/v valproic acid solution by assuming 100 mL of solution:

$M=\frac{0.10g}{100mL}*\frac{1mol}{144.21g} *\frac{1000mL}{1L}\\ \\M=6.934x10^{-3}M$

Moreover, the equilibrium expression for the valproic acid (a weak one) is written as follows:

$Ka=\frac{[H^+][A^-]}{[HA]}$

Whereas HA represents the valproic acid and A⁻ its conjugate base. Thus, by computing Ka via its pKa and writing the aforementioned equilibrium expression in terms of $x$ we obtain:

$10^{-4.8}=1.585x10^{-5}=\frac{x*x}{6.934x10^{-3}-x}$

Thus, solving for $x$, which also equals the concentration hydrogen ions, we obtain:

$x=[H^+]=6.93399x10^{-3}M$

Therefore, the pH is:

$pH=-log([H^+])=-log(6.93399x10^{-3})\\\\pH=2.16$

Best regards.