Concentration and pH Caculation of Titrations

[Squall]

Homework Statement

Calculate the volume of 0.116 M HCl(aq) required to neutralize the following solutions.
all the hydroxide ions in 25.0 mL of 0.215 M KOH(aq)

1. What is the molarity of Cl - ions at the stoichiometric point?

Homework Equations

HCl + KOH --> KCl + H2O

The Attempt at a Solution

1. So I am running out of ideas and need some help please take a look.

so at the stoichiometric point the amount of acid and base moles are equal
so .215M*.025L=.005375 moles of base that needs to be neutralized
so .005375n/.116M=.0463L

My theory is that the amount of Cl- ion in the solution is equal to the amount of moles of acid used since KCl is formed in the reaction and KCl dissociates into K+ and Cl- ions completely.

So then the molarity of Cl- at the stoichiometric point is
.005375 moles/(.0463L + .025L)=.0754M


2. Calculate the pH of the solution after the addition of 40.0 mL of 0.116 M HCl(aq) to 25.0 mL of 0.215 M KOH(aq).

.04L*.116M=.00464moles
.025L*.215M=.005375moles

so we subtract .005375n-.00464n=0.000735 moles of KOH left
no we get concentration by dividing .000735moles/(.04+.025)L=.01131 Molar
to get pH we first calculate the pOH since we still have base present
-log(.01131)=1.9466
pH= 14 - 1.9466 = 12.0533
I don't understand why this is the wrong answer seems like I did all the other problems the same way and got a right answer, but here I don't get it. If you guys can please take a look and see if I am missing something I would greatly appreciate your help.
Than You

 

[Borek]

You are doing great and these are correct answers.