Answer to Question #98911 in General Chemistry for fatty acid

Question

What volume of 0.0100 M NaOH must be added to 1.00 L of 0.0500 M HOCl to achieve a pH of 8.00? Ka for HOCl is 3.5 ́ 10–8.

Answer 1

Solution.

$HClO + OH^- = H2O + ClO^-$

$n(HClO(init)) = 0.05 \times 1.0 = 0.05 \ mol$

$n(NaOH(init)) = 0 \ mol$

$n(ClO^-(init)) = 0 \ mol$

Once we have added a certain amount of alkali, the amounts of substances will be as follows:

$n(HClO(fin)) = 0.05 - x \ mol$

$n(NaOH(fin)) = x \ mol$

$n(ClO^-(fin)) = x \ mol$

Let us use the Henderson-hasselbach equation:

$pH = pKa + lg(\frac{[ClO^-]}{[HClO]})$

$pKa = -lg(Ka) = 7.46$

$8.00 = 7.46 + lg(\frac{x}{0.05-x})$

$3.47 = \frac{x}{0.05-x}$

x = 0.039 mol

$C = \frac{n}{V}$

$V = \frac{n}{C}$

$V = \frac{0.039}{0.01} = 3.9 \ L$

Answer:

V(NaOH) = 3.9 L