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3 July, 15:06

When 25.0 mL of sulfuric acid solution was completely neutralized in a titration with 0.05 M NaOH solution, it took 18.3 mL of the NaOH (aq) to complete the job. The reaction is:

NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + H2O (l)

What was the molarity of the sulfuric acid solution?

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  1. 3 July, 15:28
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    Balanced equation will be as follows

    2NaOH + H2SO4=Na2SO4 + 2H2O

    from MaVa/MbVb=Na/Nb where Ma is molarity of acid=?

    Va is volume of acid=25.0 mL

    Mb is molarity of base = 0.05 M

    Vb is volume of base=18.3 mL

    Nb number of moles of base = 2

    Na is number of moles of acid = 1

    Ma*25 mL / 18.3mL*0.05M = 1/2

    Ma=18.3*0.05M / 50 = 0.0183M

    The molarity of sulphuric acid = 0.0183M
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