Question

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A 25.0-mL sample of H2SO4 is neutralized by 27.4 mL of
1.00M KOH. What is the concentration of the acid?

Answer 1

To calculate concentration we use -

`\:\:\:\:\:\:\:\:\:\:\:\star\longrightarrow \sf \underline{C=(n)/(V)}\\`

`\:\:\:\:\:\:\:\:\:\:\:\star\longrightarrow \sf \underline{n = C\:V}\\`

Where -

• C is the molar concentration
• n is the number of moles
• V is the volume of the solution

We are given the volume and the concentration of the KOH. Using those information, we can calculate the moles of KOH.

Given data:-

Volume of KOH, V= 27.4mL = 27.4×10⁻³ L

`\star`Concentration of KOH, C= 1 M

`\:\:\:\:\:\:\longrightarrow \sf Moles \:of \:KOH = C\:V\\`

`\:\:\:\:\:\:\longrightarrow \sf Moles \:of \:KOH = 1* 27.4 ×10⁻³\\`

`\:\:\:\:\:\:\longrightarrow \sf Moles \: of \: KOH = 0.0274  \\`

The neutralization reaction is expressed as:-

`\star\longrightarrow \sf\underline{ \pink{2KOH} + \pink{H_2SO_4} = K_2SO_4 + 2H_2O}\\`

According to this reaction, 2 moles of KOH reacted with 1 mole of H₂SO₄.Therefore, 0.0274 mole of KOH would also react with (0.0274/2)=0.0137 mole of H₂SO₄.

`\:\:\:\:\:\:\longrightarrow \sf Concentration\: of\: H₂SO₄ =\frac {Moles\:of\:H₂SO₄}{ Volume \: of \: H₂SO₄}\\`

`\:\:\:\:\:\:\longrightarrow \sf Concentration\: of\: H₂SO₄ =\frac {0.0137}{ 25×10⁻³  }\\`

`\pink{\because\sf \underline{ Volume\: of \: H₂SO₄= 25 mL = 25×10⁻³ L}}\\`

`\:\:\:\:\:\:\longrightarrow \sf Concentration\: of\: H₂SO₄ =\frac {0.0137}{ 0.025 }\\`

`\:\:\:\:\:\:\longrightarrow \sf \underline{Concentration\: of\: H₂SO₄ = 0.548\: M}\\`

Therefore, the concentration of H₂SO₄ is 0.548M.