How many mL of .250 M NaOh would be needed to neutralize a 25 mL sample of the sulfuric acid prepared in part a?

Question

A- Calculate the number of grams of sulfuric acid at the must be used to produce exactly 500 of a 0.345 M solution of sulfuric acid

answer- 16.56 grams H2SO4

Answer 1

I’ll assume the question is asking for exactly 500 mls of 0.345 M H2SO4.

molar mass H2SO4 = 98.08 g/mol

0.345 M = 0.345 mol/L

To make 500 mls (0.500 L) you need 0.345 mol/L x 0.500 L = 0.1725 moles

0.1725 mols x 98.08 g/mol = 16.92 g (not sure where you got 16.56 g)

To answer the part about NaOH, look at the balanced equation

2NaOH + H2SO4 ==> Na2SO4 + 2H2O

moles H2SO4 = 25 ml x 1 L/1000 ml x 0.345 mol/L = 0.008625 mols

mols NaOH needed = 0.008625 mol H2SO4 x 2 mol NaOH/ mol H2SO4 = 0.01725 mols

mls NaOH = 0.01725 mols x 1 L / 0.250 mols = 0.069 L = 69 mls NaOH needed