Question

If 15.00 mL of bleach (contains HOCl) requires 45.2 mL of 0.200 M KOH to reach the equivalence point, calculate the number of grams of HOCl contained in a 1.0 L bottle of bleach. Show all steps.

Answer 1

• At the equivalence point equality is observed for reacting substances: $C(1)⋅V(1)=C(2)⋅V(2)$ , in particular for this question: $C(HClO)⋅V(HClO)=C(KOH)⋅V(KOH)$. On the basis of this equality, it`s necessary to find C(HClO):

C(HClO) = $(C(KOH)⋅V(KOH))/V(HClO)$ = (0.2(M) ⋅ 0.05(L))/0.015(L) = 0.67 M

• The next step is to find the amount of substance of HClO with concentration 0.67M in 1 L:

$\nu(HClO) = C(HClO) ⋅ V(HClO)$ = 0.67(M) ⋅ 1(L) = 0.67 mol

• Now, we can find the mass of HClO in 1 L:

$m(HClO) = \nu ⋅ M$ = 0.67(mol) ⋅ 52.46(g/mol) = 35.15 g

Answer: m(HClO)=35.15 g.