I'm trying to find the molarity of H2SO4 in this calculation. I attempted it but I still need someone to check my work. I used 10.00 mL of H2SO4 and the volume added to it was 24.97 mL of the 0.0924M NaOH solution. The equation is 2NaOH + H2SO4 - > 2H2O + Na2SO4, so I'm aware 1 mol of H2SO4 is 2 mol NaOH.

What I tried: 0.0924 mol/L x 0.02497 L = 0.002307 mol NaOH

x 1 mol H2SO4/2 mol NaOH = 0.001156 mol H2SO4

0.0011536 mol H2SO4/0.01000 L H2SO4 = 0.1154M H2SO4

Am I doing this right?

Question

Lizbeth Solomon

Chemistry

22 September, 02:06

I'm trying to find the molarity of H2SO4 in this calculation. I attempted it but I still need someone to check my work. I used 10.00 mL of H2SO4 and the volume added to it was 24.97 mL of the 0.0924M NaOH solution. The equation is 2NaOH + H2SO4 - > 2H2O + Na2SO4, so I'm aware 1 mol of H2SO4 is 2 mol NaOH.

What I tried: 0.0924 mol/L x 0.02497 L = 0.002307 mol NaOH

x 1 mol H2SO4/2 mol NaOH = 0.001156 mol H2SO4

0.0011536 mol H2SO4/0.01000 L H2SO4 = 0.1154M H2SO4

Am I doing this right?

+3

Answers (1)

Know the Answer?

Azaria Cabrera · Answer 1

Since it goes to completion, we can use:

Use NaMaVa = NbMbVb

where

Na = moles of H + in H2SO4

Ma = molarity of the acid

Va = volume of the acid

Nb = moles of OH - in NaOH

Mb = molarity of the base

Vb = molarity of the base

plug in the numbers

(2 moles of H+) (Volume acid) (10.00mL acid) = (1 mole of OH-) (.248M base) (18.71mL base)

20 V = 4.64

V=.232 M