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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?

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Answers (1)
  1. 22 September, 02:28
    0
    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
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