Chemistry Forum

Hi everyone, I have the following exercise:
"A solution containing 1.15 g of a mixture of NaOH and KOH is neutralised with 48.4 mL of 0.500 N H₂SO₄ solution. Calculate the percentages by weight of NaOH and KOH in the original mixture."
Would you help me to understand how to solve it? Thank you

two equations

m(NaOH) + m(KOH) = 1.15 g

n(NaOH) +n(KOH) = 48.4 ml * 0.5 mmol/ml

use n =m/M

thanks, I've already solved it