Answer :
Complete Question
The complete question is shown on the first uploaded image
Answer:
The pH is [tex]pH = 4.94[/tex]
Explanation:
From the question we are told that
The average concentration of NaOH is [tex][NaOH] = 0.101 M[/tex]
The volume of NaOH is [tex]V__{NaOH}} = 15.00 mL[/tex]
The average concentration of Acetic acid is [tex][Acetic \ Acid] =0.497 \ M[/tex]
The volume of Acetic acid is [tex]V__{Acetic \ Acid}} = 5.00 \ mL[/tex]
The chemical equation for this reaction is
[tex]NaOH + CH_3COOH ---> CH_3 COONa + H_2 O[/tex]
The total volume of the solution is
[tex]V__{Total}} = V__{NaOH}} + V__{Acetic \ Acid}}[/tex]
Substituting values
[tex]V__{Total}} = 15 + 5[/tex]
[tex]V__{Total}} = 20mL = 20 *10^{-3} L[/tex]
The number of moles of NaOH is mathematically represented as
[tex]n__{NaOH}} = [NaOH] * V__{NaOH}}[/tex]
substituting values
[tex]n__{NaOH}} = 0.101 * 15*10^{-3}[/tex]
[tex]n__{NaOH}} = 0.001515 \ moles[/tex]
The number of moles of Acetic acid is mathematically represented as
[tex]n__{Acetic acid}} = [Acetic \ acid] * V__{Acetic acid}}[/tex]
substituting values
[tex]n__{Acetic acid}} = 0.497 * 5*10^{-3}[/tex]
[tex]n__{Acetic acid}} = 0.002485\ moles[/tex]
From the chemical equation
1 mole of NaOH reacts with 1 mole of Acetic acid to produce 1 mole of [tex]CH_3 COONa[/tex] salt and 1 mole of [tex]H_2 O[/tex]
So
0.001515 moles of NaOH reacts with 0.001515 moles of Acetic acid to produce 0.001515 moles of [tex]CH_3 COONa[/tex] salt and 0.001515 moles of [tex]H_2 O[/tex]
This implies the number of moles of NaOH remaining after the react would be
[tex]\Delta n__{NaOH}} = 0.001515 - 0.001515[/tex]
[tex]\Delta n__{NaOH}} = 0 \ mole[/tex]
the number of moles of Acetic acid remaining after the react would be
[tex]\Delta n__{Acetic acid}} = 0.002485 - 0.001515[/tex]
[tex]\Delta n__{Acetic acid}} = 0.00097 \ moles[/tex]
the number of moles of [tex]CH_3 COONa \ salt[/tex] remaining after the react would be
[tex]\Delta n__{CH_3 COONa \ salt}} = 0 + 0.001515[/tex]
[tex]\Delta n__{CH_3 COONa \ salt}} = 0.001515 \ moles[/tex]
the number of moles of [tex]H_2 O[/tex] remaining after the react would be
[tex]\Delta n__{H_2O}} = 0 + 0.001515[/tex]
[tex]\Delta n__{H_2O}} = 0.001515 \ moles[/tex]
The expected pH is mathematically evaluated as
[tex]pH = pK_a + log [\frac{[CH_3 COONa]}{[Acetic \ acid]} ][/tex]
Where [tex]pKa[/tex] is mathematically evaluated as
[tex]pK_a = - log (K_a)[/tex]
The concentration of [tex]CH_3 COONa \ salt[/tex] is mathematically evaluated a s
[tex][CH_3 COONa] = \frac{\Delta n_CH_3 COONa \ salt }{V__{Total}}}[/tex]
substituting values
[tex][CH_3 COONa] = \frac{0.001515}{20 *10^{-3}}[/tex]
[tex][CH_3 COONa] = 0.07575M[/tex]
The concentration of Acetic acid is mathematically evaluated as
[tex][Acetic acid] = \frac{\Delta n__Acetic acid}{V__{Total}}}[/tex]
substituting values
[tex][CH_3 COONa] = \frac{0.00097}{20 *10^{-3}}[/tex]
[tex][CH_3 COONa] = 0.0485 M[/tex]
Substituting values into the equation for pH
[tex]pH = - log (1 .8 *10^{-5}) + log [\frac{0.07575}{ 0.0485} ][/tex]
[tex]pH = log [\frac{0.07575}{ 0.0485} ] - log (1 .8 *10^{-5})[/tex]
[tex]pH = log [\frac{1.561856}{1.8*10^{-5}} ][/tex]
[tex]pH = 4.94[/tex]
