Calculate pH? Lookin for the best answer!?

Question

Calculate pH:

25mL of .2 M CH3OOH + 10 mL of .2 M NaOH? Ka= 1.8E-5

Step-by-step procedure please! I have to do 20 of these!

Answer 1

Moles of acid:

0.025L * 0.2mole/L = 0.005 moles

Moles of NaOH:

0.01L * 0.2mole/L = 0.002moles

Concentration of acid:

0.005mol/0.030L = 0.17M

Concentration of NaOH:

0.002mol/0.030L = 0.07

Ka= [CH3COO-][H+]/ [CH3OOH]

1.8E-5 = X^2/0.17

X= 0.00175M = [H+]

*NaOH dissociates completely there for [NaOH] = [Na+] = [OH-] = 0.07M

In solution:

[OH-] - [H+] =  0.068[OH-]

pOH = -log(0.068) = 1.17

pH = 14 - pOH = 12.83

Answer 2

Given that Ka is 1.8e-5, all u have to do i

-log(Ka) = -log(1.8e-5) =  4.74472749...

So the pH is around 4.74