Exactly 17.0 mL of a H2SO4 solution was required to neutralize 45.0 mL of 0.235 M NaOH. What was the concentration of the H2SO4 solution? Given: H2SO4 (aq) + 2NaOH (aq) → 2H2O (l) + Na2SO4 (aq) Exactly 17.0 mL of a H2SO4 solution was required to neutralize 45.0 mL of 0.235 M NaOH. What was the concentration of the H2SO4 solution? Given: H2SO4 (aq) + 2NaOH (aq) → 2H2O (l) + Na2SO4 (aq) 0.311 M 0.622 M 5.63 M 0.00529.

Question

Chemistry

Joy Curry

6 November, 04:36

Exactly 17.0 mL of a H2SO4 solution was required to neutralize 45.0 mL of 0.235 M NaOH. What was the concentration of the H2SO4 solution? Given: H2SO4 (aq) + 2NaOH (aq) → 2H2O (l) + Na2SO4 (aq) Exactly 17.0 mL of a H2SO4 solution was required to neutralize 45.0 mL of 0.235 M NaOH. What was the concentration of the H2SO4 solution? Given: H2SO4 (aq) + 2NaOH (aq) → 2H2O (l) + Na2SO4 (aq) 0.311 M 0.622 M 5.63 M 0.00529.

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Answers (1)

Know the Answer?

Kamila Jackson · Answer 1

Molarity for the sulfuric acid is 0.622 M

Explanation:

When we neutralize an acid with a base, molarity of both. both volume are the same. The formula is:

M acid. volume of acid = M base. volume of base

M acid = unknown

Volume of acid = 17 mL

Volume of base = 45 mL

M base = 0.235 M

Therefore, we replace: M acid. 17 mL = 0.235 M. 45 mL

M acid = (0.235 M. 45 mL) / 17 mL

M acid = 0.622 M