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If 42.68 ml of 0.43 M KOH is used to neutralize 18.40 ml of H, SO4, calculate

the concentration of the sulfuric acid solution.
Write the net ionic equation of this reaction

User Mauro
by
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1 Answer

7 votes

The concentration of the sulfuric acid : 0.499 M

The net ionic equation

2 OH ⁻(aq]+ 2 H ⁺ (aq] → 2 H ₂O (l]

Further explanation

Given

42.68 ml of 0.43 M KOH

18.40 ml of H2SO4

Required

the concentration of the sulfuric acid

the net ionic equation

Solution

Acid-base titration formula

Ma. Va. na = Mb. Vb. nb

Ma, Mb = acid base concentration

Va, Vb = acid base volume

na, nb = acid base valence

Input the value :

a=KOH, b= H2SO4

0.43 x 42.68 x 1 = Mb x 18.40 x 2

Mb = 0.499 M

The net ionic equation

Reaction

2KOH + H2SO4 → K2SO4 + 2H2O

2 K ⁺ (aq]+ 2 OH ⁻ (aq] + 2 H ⁺ (aq] + SO ₄²⁻(aq]2 K ⁺ (aq] + SO ₄²⁻(aq] + 2 H ₂ O (l]

canceled the spectator ions :

2 OH ⁻(aq]+ 2 H ⁺ (aq] → 2 H ₂O (l]

User Matt Waldron
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