Answer:
8.73
Explanation:
The concentration of acetic acid can be determined as follows:
[tex]M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)[/tex]
[tex]M_{KOH}=0.113 M\\V_{KOH}=79.06 mL[/tex]
[tex]V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????[/tex]
[tex]M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}[/tex]
[tex]M_{CH3COOH} = \frac{0.113*79.06}{95.27}[/tex]
[tex]M_{CH3COOH} = 0.094 M[/tex]
Moles of [tex]CH_3COOH[/tex] = [tex]95.27* 10^{-3}* 0.094[/tex]
=0.0090 moles
Moles of [tex]KOH = 79.06*10^{-3}*0.113[/tex]
= 0.0090 moles
The equation for the reaction can be expressed as :
[tex]CH_3COOH[/tex] [tex]+[/tex] [tex]KOH[/tex] -----> [tex]CH_3COO^{-}K^+[/tex] [tex]+[/tex] [tex]H_2O[/tex]
Concentration of [tex]CH_3COO^{-[/tex] ion = [tex]\frac{0.0090}{Total volume (L)}[/tex]
= [tex]\frac{0.0090}{(95.27+79.06)} *1000[/tex]
= 0.052 M
Hydrolysis of [tex]CH_3COO^{-[/tex] ion:
[tex]CH_3COO^{-[/tex] [tex]+[/tex] [tex]H_2O[/tex] -----> [tex]CH_3COOH[/tex] [tex]+[/tex] [tex]OH^-[/tex]
[tex]K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}[/tex]
⇒ [tex]\frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}[/tex]
= [tex]0.5494*10^{-9}= \frac{x*x}{0.052-x}[/tex]
As K is so less, then x appears to be a very infinitesimal small number
0.052-x ≅ x
[tex]0.5494*10^{-9}= \frac{x^2}{0.052}[/tex]
[tex]x^2 = 0.5494*10^{-9}*0.052[/tex]
[tex]x^2 = 0.286*10^{-10[/tex]
[tex]x = \sqrt{0.286*10^{-10[/tex]
[tex]x =0.535*10^{-5}M[/tex]
[tex][OH] = x =0.535*10^{-5}[/tex]
[tex]pOH = -log[OH^-][/tex]
[tex]pOH = -log[0.535*10^{-5}][/tex]
[tex]pOH = 5.27[/tex]
pH = 14 - pOH
pH = 14 - 5.27
pH = 8.73
Hence, the pH of the titration mixture = 8.73
