# Answer to Question #33187 in Inorganic Chemistry for Larry

Question #33187

What is the molarity of a sodium hydroxide solution if 27.9 mL of this solution reacts exactly with 22.30 mL of 0.253 M sulfuric acid?

Expert's answer

n(H

n(H

The balanced equation is:

H

the ratio between H

n(NaOH) = 0.00564/2 = 0.00282 mol

C

C

Answer: 0.113 M

_{2}SO_{4}) = C_{m}x Vn(H

_{2}SO_{4}) = 0.253 M x 0.02230 L = 0.00564 molThe balanced equation is:

H

_{2}SO_{4}+ 2NaOH = Na_{2}SO_{4}+ 2H_{2}Othe ratio between H

_{2}SO_{4}and NaOH is 1 : 2n(NaOH) = 0.00564/2 = 0.00282 mol

C

_{m}(NaOH) = n(NaOH)/V(NaOH)C

_{m}(NaOH) = 0.00282 mol/ 0.0250 L = 0.113 MAnswer: 0.113 M

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