What volume (in mLs) of 0.100 M sodium hydroxide solution must be added to 20.0 mL of 0.13 M sulfuric acid ..?

Hi! just wondering, What volume (in mLs) of 0.100 M sodium hydroxide solution must be added to 20.0 mL of 0.13 M sulfuric acid solution to reach the end point?

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  • Dr.A
    Lv 7
    10 years ago
    Favorite Answer

    moles H2SO4 = 0.13 M x 0.0200 L=0.0026

    H2SO4 + 2 NaOH = Na2SO4 + 2 H2O

    moles NaOH required = 0.0026 x 2 =0.0052

    V = 0.0052/ 0.100 L=0.052 L => 52 .0 mL

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  • 3 years ago

    to make certain the staggering volume of the three M answer use M1*V1 = M2*V2 the place: M1 = molarity of first answer V1 = volume of first answer M2 = molarity of 2nd answer V2 = volume of 2nd answer the 1st answer right here may be the ten mL of 18 M H2SO4 and the 2nd answer may be the three M answer (18 mol/L)*(0,01 L) = (3 mol/L)*(V2 L) remedy for V2 (18 mol/L)*(0.01 L)/(3 mol/L) = V2 L V2 = 0.06 L or the staggering volume of the diluted answer is 60 mL on account that there have been 10 mL to start up, to get to a 3M answer, 50 mL of water desires to be further. As a know protection - by no potential upload water to centred stable acid. upload the acid to the water.

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