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 [tex]CH_3COOH[/tex] and [tex]NaOH[/tex].
[tex]\text{Moles of }CH_3COOH=\text{Concentration of }CH_3COOH\times \text{Volume of solution}=0.35M\times 0.052L=0.018mole[/tex]
[tex]\text{Moles of }NaOH=\text{Concentration of }NaOH\times \text{Volume of solution}=0.40M\times 0.027L=0.011mole[/tex]
The balanced chemical reaction is,
[tex]CH_3COOH+OH^-\rightarrow CH_3COO^-+H_2O[/tex]
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.
[tex][H^+]=\frac{\text{Moles of }H^+}{\text{Total volume}}[/tex]
[tex][H^+]=\frac{0.0007mole}{(52+27)mL}=8.9\times 10^{-6}mole/mL=8.9\times 10^{-3}M[/tex]
Now we have to calculate the pH.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (8.9\times 10^{-3})[/tex]
[tex]pH=2.05[/tex]
Therefore, the pH after 27.0 mL of NaOH added will be 2.05
