Calculate the pH of the solution that results when 27.38mL of 0.15M H2SO4(aq) is mixed with 18.50mL of 0.19M Ca(OH)2 (aq)

Answer 1

mL x M = millimoles.
27.38 x 0.15 = 4.197 mmoles H2SO4
18.5 x 0.19 = 3.515 mmoles Ca(OH)2
(Watch the significant figures. I have used too many assuming you did not omit zeros on the 0.15 and 0.19).

......H2SO4 + Ca(OH)2 ==> CaSO4 + 2H2O
init...4.107....3.515........0....0
change.-3.515...3.515....3.515.3.515
equil...0.592....0........3.515..3.515

So the ICE chart tells you that you have an excess of 0.592 mmoles H2SO4 in (27.38+18.5 mL) for a M = 0.592 moles/45.88 mL = ?M
Determine the pH of that solution
Remember H2SO4 has a k2 value of about 0.012.

Answer 2

0.012