Find the volume of H2SO4. Find the volume of NaOH. Find the concentration of H2SO4.

this question should to be in 20 points...anyway hope it help..

all is in the explanation.

ans; 0.044

Explanation: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

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