Determine the volume of 0.225 M KOH solution required to neutralize each sample of sulfuric acid. The neutralization reaction is:
H2SO4(aq) + 2 KOH(aq) → K2SO4(aq) + 2H2O(l)
(a) 45 mL of 0.225 M H2SO4
(b) 185 mL of 0.125 M H2SO4
(c) 75 mL of 0.100 M H2SO4
Solution: Here, we are going to calculate the volume of KOH solution required to neutralize the given H2SO4 solutions.
Step1:
Neutralisation reaction is the reaction between an acid and a base to give salt and water. The neutralization reaction in the given problem is:
2KOH(aq) + H2SO4(aq) → 2H2O(l) + K2SO4(aq)
Here, 1 mole of H2SO4 requires 2 moles of KOH for complete neutralization.
Step2:
Again, molarity(M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.
Moles of solute
Molarity = ------------------------------- ------(1)
Volume of solution in litre
Step3:
Given, volume of H2SO4 solution = 45 mL = 45/1000 L = 0.045 L [1000 mL = 1 L]Molarity of H2SO4 solution = 0.225 M
Therefore, number of moles of H2SO4 = Molarity x volume of solution in litres
= 0.225 M x 0.045 L
= 0.010125 mol
Step4:
Number of moles of KOH = 2 x number of moles of H2SO4 = 2 x 0.010125 mol = 0.02025 mol
Molarity of KOH solution = 0.225 M
Therefore, volume of KOH solution required = moles of KOH / molarity
= 0.02025 mol / 0.225 M
= 0.09 L
= 0.09 x 1000 mL [1 L = 1000 mL]
= 90 mL
Thus, 90 mL of KOH solution will be required.
Step5:
b) Given, volume of H2SO4 solution = 185 mL = 185/1000 L = 0.185 L [1000 mL = 1 L]
Molarity...
