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:
[tex]M=\frac{0.10g}{100mL}*\frac{1mol}{144.21g} *\frac{1000mL}{1L}\\ \\M=6.934x10^{-3}M[/tex]
Moreover, the equilibrium expression for the valproic acid (a weak one) is written as follows:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
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 [tex]x[/tex] we obtain:
[tex]10^{-4.8}=1.585x10^{-5}=\frac{x*x}{6.934x10^{-3}-x}[/tex]
Thus, solving for [tex]x[/tex], which also equals the concentration hydrogen ions, we obtain:
[tex]x=[H^+]=6.93399x10^{-3}M[/tex]
Therefore, the pH is:
[tex]pH=-log([H^+])=-log(6.93399x10^{-3})\\\\pH=2.16[/tex]
Best regards.
