I'm doing an acids and bases question, the question is:
At 25 °C, the acid dissociation constant Ka for ethanoic acid has the value
1.75 × 10–5 mol dm–3.
Calculate the pH of the solution formed when 10.0 cm3 of 0.154 mol dm–3 potassium
hydroxide are added to 20.0 cm3 of 0.154 mol dm–3 ethanoic acid at 25 °C.
I have calculated the moles of KOH and CH3COOH and calculated the excess KOH (3.08*10-3)
According to the mark scheme, I must times this figure by (1000/60), but I don't know why? I thought it would be the other way round to get the total volume in dm3?
Thanks.
[EDIT] Now understand and feel very stupid