WHAT is the resulting PH if 400ML of 0.50M HCL is mixed with 300ML of 0.080M NAOH?

Question

1 0.4
2. 1.3
3 0.6
4 7.0
5 2005
6 4.5

Answer 1

Assume there is a state after complete mixing but before reaction.
HCl concentration after mixing: 400*0.50M/(400+300) = 0.286M
NaOH concentration after mixing: 300*0.080M /(400+300) = 0.0343M
Thus NaOH is the limiting reactant. After reaction of:
HCl + NaOH ==> NaCl + H2O,
we have HCl left: 0.286M - 0.0343M = 0.251M, that means [H+] = 0.251M. Therefore:
pH = -log([H+]) = -log(0.251) = 0.60