Question

HELP!! 1.20g of sodium hydrogen sulphate was made into 100cm3 standard solution. 25cm3 of this solution was neutralised by 23.55cm3 NaOH. What was the concentraion of NaOH?

Answer 1

NaHSO4 + NaOH --> Na2SO4 + H2O
mols NaHSO4 = grams/molar mass = 1.20/ 120 = 0.01
M NaHSO4 = mols/L = 0.01/0.1 = 0.1 M
mols NaHSO4 used = M x L = 0.1M x 0.025 = 0.0025
Since 1 mol NaHSO4 = 1 mol NaOH from the balanced equation, then mols NaOH = 0.0025 and M = mols NaOH/L NaOH = 0.0025/0.02355 = 0.1062
Personally, I dislike all those zeros. It makes it easy to loose one or more of them. I prefer millimoles like this.
millimols NaHSO4 used = mL x M = 25 x 0.1 M = 2.5
millimols NaOH = 2.5
M NaOH = millimoles/mL = 2.5/23.55 = 0.1062 M