14.1 mL H₂SO₄ is needed to neutralize solution of NaOH.
Balanced chemical equation for neutralization reaction of sulfuric acid and sodium hydroxide:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
c(H₂SO₄) = 0.3 M = 0.3 mol/L; concentration of sulfuric acid
V(NaOH) = 34.0 mL = 0.034 L; volume of sodium hydroxide
c(NaOH) = 0.25 M = 0.25 mol/L; concentration of sodium hydroxide
n(NaOH) = c(NaOH) × V(NaOH)
n(NaOH) = 0.25 mol/L × 0.034 L.
n(NaOH) = 0.0085 mol; amount of sodium hydroxide
From chemical reaction: n(H₂SO₄) : n(NaOH) = 1 : 2.
n(H₂SO₄) = 0.0085 mol ÷ 2
n(H₂SO₄) = 0.00425 mol; amount of sulfuric acid
V(H₂SO₄) = n(H₂SO₄) ÷ c(H₂SO₄).
V(H₂SO₄) = 0.00425 mol ÷ 0.3 mol/L.
V(H₂SO₄) = 0.0141 L = 14.1 mL; volume of sulfuric acid
More about neutralization: brainly.com/question/23008798
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