J.R. S. answered • 04/26/17

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So, this is, or can be, a rather complicated calculation depending on the assumptions, etc. However, in its simplest form..

2NaOH + H2SO4 ==> Na2SO4 + 2H2O

Assuming a concentration of H2SO4 of 18.4 M (typical for concentrated H2SO4), first determine volume of H2SO4 needed to completely neutralize 10 ml of 0.1 M NaOH:

- 0.010 L x 0.1 mol/L = 1x10
^{-3}moles NaOH present - 1x10
^{-3}moles NaOH x 1 mole H2SO4/2 moles NaOH = 5x10^{-4}moles H2SO4 for complete neutralization (pH = 7) - (x L)(18.4 mol/L) = 5x10
^{-4}moles and x = 2.7x10^{-5}L of H2SO4 needed for complete neutralization (ph = 7)

Next, determine volume H2SO4 to go from pH 7 to pH 1.5

- Assume that the volume (2.7x10^-5 L or 27 ul) did not change the volume of the 10 mls so you still have 10 ml
- pH 1.5 means [H
^{+}] = 3.2x10^{-2}M^{} - (x L)(18.4 M) = (0.010 L)(3.2x10
^{-2}M) and x = 1.7x10^{-5}L (17 ul) H2SO4 needed. This also assumes that only the first ionization of H2SO4 is considered in supplying the H^{+}

Thus, the total volume of 18.4 M H2SO4 needed to decrease the pH of 10 ml of 0.1 M NaOH to 1.5 would be 27 ul + 17 ul = 44 ul = 0.044 mls