The more I think about it the wronggrrrr I get!!! what am I missing here?

Question

Two solutions are made. in the first beaker 0.5g of HCl is dissolved into 3.5L of water, in the second beaker 1.5g of KOH is dissolved in 6L of water - the two solutions are then poured into a large bucket. Determine the pH of the resulting liquid

Answer 1

0.5 g HCl * (1 mol HCl / 36.46 g HCl) = 0.0137 mol HCl

1.5 g KOH * (1 mol NaOH / 56.1056 g KOH) = 0.0267 mol NaOH

they react 1 : 1 (HCl + KOH = NaCl + H2O) so the limiting reagent is the HCl, leaving you with (0.0267- 0.0137) = 0.0130 moles of KOH after neutralization.

The total volume after the 2 solutions are mixed is 9.5 L. so the concentration of OH- is 0.0130 mol / 9.5 L = 1.36 * 10 ^-3 M.

plugging that into the pOH = -log [OH] gives us 2.863 so the pH is equal to 14 - 2.863, which is 11.13.

Let me know if you have any questions about the procedure. Hope this helps.