What is the molarity of a sodium hydroxide solution if 35.4 ml of this solution is neutralized by 24.2 ml of 1.19 m sulfuric acid solution?

ANSWERS

2016-04-14 23:01:32

The item above can be calculated by equating the equivalents of the solutions. The equivalent of the solution is calculated by through normality. Normality is equal to molarity if the compound has only 1 equivalent. 1.19 M H2SO4 = 2.38 N H2SO4 N1V1 = N2V2 For sodium hydroxide, NaOH, molarity is equal to normality. (N1)(35.4 mL) = (2.38)(24.2 mL) N1 = 1.627 N = 1.627 M Thus, the molarity of the NaOH solution is equal to 1.627.

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