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[Grade 11] Could I get some help with this easy titration question?
'A student used 0.118mol/L H2(SO4) to titrate 10.00ml of Na(OH). The initial volume reading of the burette was 1.05ml The final volume reading on the burette was 23.00ml What is the concentration of the Na(OH), according to this data?'
The back of the sheet says the answer is 0.518. I would really appreciate somebody to walk me through this question.
2 Answers
- electron1Lv 74 years agoFavorite Answer
To determine the number of moles, multiply the volume in liters by the molarity. Since sulfuric acid has two hydrogen atoms, one mole of sulfuric acid will neutralize two moles of sodium hydroxide. To determine the number of moles of sulfuric acid, multiply the volume in liters by the concentration. To determine the volume of H2SO4, subtract 1.05 ml from 23.00 ml.
23 – 1.05 = 21.95 ml = 0.02195 L
n = 0.02195 * 0.118 = 0.0025901
For NaOH, n = 2 * 0.0025901 = 0.00151802
Let’s convert the volume of sodium hydroxide to liters.
V = 0.01 L
0.01 * M = 0.00151802
M = 0.00151802 ÷ 0.01 = 0.51802
This rounds to 0.518. I hope this helps you to understand how to solve this type of problem.
- Roger the MoleLv 74 years ago
H2SO4 + 2 NaOH → Na2SO4 + 2 H2O
(23.00 - 1.05) mL x (0.118 mol/L H2SO4) x (2 mol NaOH / 1 mol H2SO4) / (10.00 mL NaOH) = 0.518 mol/L
