Asked by BOB
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?
Answers
Answered by
DrBob222
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
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
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