How many mL of a 10.4% solution by mass of KOH would you need to dilute to a final volume of 15 L to achieve a pH equal to 11.15? The molecular weight of KOH is 56.1 g/mol...
attempt at a solution:
(15L)(10^(-2.85)mol/L OH-)(1:1 mol ratio) = 0.02119 mol of KOH
0.02119 mol KOH (56.1 g/mol)(100 g solution/10.4 g KOH).
but I don't know the density of the solution so I don't know how to complete the last step.. I could be completely wrong in my methods but this is what makes sense to me..
any help on how to solve the problem would be appreciated!!!