Question

A volume of 90.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 21.2 mL of 1.50 M H2SO4 was needed? The equation is

2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

Express your answer with the appropriate units.

Answer 1

The answer to your question: 0.7 M

Step-by-step explanation:

Data

V of KOH = 90 ml

[KOH] = ?

V H2SO4 = 21.2 ml

[H2SO4] = 1.5 M

2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)

Molarity = moles / volume

moles of H₂SO₄ = (1.5) (21.2)

= 31.8

2 moles of KOH -------------- 1 mol of H₂SO₄

x -------------- 31.8 mol of H₂SO₄

x = (31.8)(2) / 1

x = 63.8 moles of KOH

Molarity = 63.8 / 90

= 0.7 M