I need help doing this on my chem postlab, please explain how u got the pH? Thank You?

Question

Please calculate the pH of each of the following aqueous solutions. Report all pH's to two decimal places.
10.0 mL of 0.1 M NaC2H3O2 plus 10.0 mL of 0.1 M NaOH

10.0 mL of 0.1 M NaC2H3O2 plus 15.0 mL of 0.1 M NaOH

Answer 1

Sodium acetate is the salt of acetic acid which is a weak acid with pKa=4.76. Thus the acetate ion is a weak base with pKb=pKw-pKa = 14-4.76 =9.24 and will hydrolyze according to the reaction :
CH3COO- +H2O <=> CH3COOH + OH-

The presence of the strong base NaOH will shift the equilibrium to the left, reducing the contribution of OH- by acetate.

First of all let's find the concentrations just after mixing.

Cacetate= mole/Vfinal =M1V1/(V1+V2) =0.1*10/(10+10)= 1/20= 0.05 M
CNaOH= M2V2/(V1+V2)= 0.1*10/20= 0.05 M

.. .. .. .. .. .. CH3COO- +H2O <=> CH3COOH + OH-
Initial .. .. .. .. .. 0.05 .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. 0.05
React .. .. .. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. .. ..x
At Equil. .. .. . 0.05-x .. .. .. .. .. .. .. .. .. .. x .. .. 0.05+x

Kb= [CH3COOH][OH-] /[CH3COO-] =>
x(0.05+x)/(0.05-x) =10^-9.24 =5.75*10^-10

Practically NaOH is so concentrated and acetate so weak that it will be as if you had only NaOH present. Anyway let's do the calculations
0.05x+x^2= 0.05*5.75*10^-10 -5.75*10^-10x=>
x^2 + 0.05x -2.875*10^-11 = 0 (5.75*10^-10x is too small compared to 0.05x so we can leave out)
x=5.75*10^-10

so pH=14-pOH= 14-(-log(0.05+x))= 14-(-log(0.05+5.75*10^-10)) =12.698 =12.70
which is exactly the same as if we ignore x and have pH=14-(-log(0.05))

Similarly in the second case
CNaOH= 0.1*15/25 =0.06 and pH =14-(-log(0.06)) =12.778 =12.78