Answer : The pH after 27.0 mL of NaOH added will be 2.05
Explanation : Given,
Concentration of acetic acid = 0.35 M
Volume of acetic acid = 52.0 mL = 0.052 L (conversion used : 1 L = 1000 mL)
Concentration of NaOH = 0.40 M
Volume of NaOH = 27.0 mL = 0.027 L
First we have to calculate the moles of
and
.


The balanced chemical reaction is,

Initial moles 0.018 0.011 0
At eqm. moles (0.018-0.011) 0 0.011
= 0.0007
Now we have to calculate the hydrogen ion concentration.
![[H^+]=\frac{\text{Moles of }H^+}{\text{Total volume}}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DH%5E%2B%7D%7B%5Ctext%7BTotal%20volume%7D%7D)
![[H^+]=\frac{0.0007mole}{(52+27)mL}=8.9\times 10^{-6}mole/mL=8.9\times 10^{-3}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5Cfrac%7B0.0007mole%7D%7B%2852%2B27%29mL%7D%3D8.9%5Ctimes%2010%5E%7B-6%7Dmole%2FmL%3D8.9%5Ctimes%2010%5E%7B-3%7DM)
Now we have to calculate the pH.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)


Therefore, the pH after 27.0 mL of NaOH added will be 2.05