Respuesta :
To determine the pH of the resulting solution, we need to determine the concentration of hydrogen ions in the solution. To do this we use, the acid dissociation constant of C6H5COOH. We do as follows:
C6H5COOH = H+ + C6H5COO-
C6H5COONa = Na+ + C6H5COO-
HCl = H+ + Cl-
Using the ICE table before adding HCl,
C6H5COOH H+ C6H5COO-
I 0.250 0 0.20
C -x x x
---------------------------------------------------------------
E 0.250 - x x 0.20+x
Ka = 6.5 x 10^-5 = x(0.20+x) / 0.250-x
x = 8.12x10^-5 M
0.20+x = 0.2000812 M
0.250 - x = 0.24992 M
After addition of HCl,
[H+] = 8.12x10^-5 M (0.250 L) + 0.100 M (.025 L) / 0.025 +0.250 = 9.165x10^-3 M
[C6H5COO-] = 0.2000812 M (.250) / .025 + .250 =0.1819 M
[C6H5COOH] = 0.24992 M ( .250 ) / .025 + .250 = 0.2272 M
C6H5COOH H+ C6H5COO-
I 0.2272 9.165x10^-3 0.1819
C -x x x
---------------------------------------------------------------
E 0.2272 - x 9.165x10^-3+x 0.1819+x
6.5 x 10^-5 =(9.165x10^-3+x )(0.1819+x) / 0.2272-x
x = - 9.08x10^-3 M
pH = -log [9.165x10^-3 - 9.08x10^-3] =1.74
C6H5COOH = H+ + C6H5COO-
C6H5COONa = Na+ + C6H5COO-
HCl = H+ + Cl-
Using the ICE table before adding HCl,
C6H5COOH H+ C6H5COO-
I 0.250 0 0.20
C -x x x
---------------------------------------------------------------
E 0.250 - x x 0.20+x
Ka = 6.5 x 10^-5 = x(0.20+x) / 0.250-x
x = 8.12x10^-5 M
0.20+x = 0.2000812 M
0.250 - x = 0.24992 M
After addition of HCl,
[H+] = 8.12x10^-5 M (0.250 L) + 0.100 M (.025 L) / 0.025 +0.250 = 9.165x10^-3 M
[C6H5COO-] = 0.2000812 M (.250) / .025 + .250 =0.1819 M
[C6H5COOH] = 0.24992 M ( .250 ) / .025 + .250 = 0.2272 M
C6H5COOH H+ C6H5COO-
I 0.2272 9.165x10^-3 0.1819
C -x x x
---------------------------------------------------------------
E 0.2272 - x 9.165x10^-3+x 0.1819+x
6.5 x 10^-5 =(9.165x10^-3+x )(0.1819+x) / 0.2272-x
x = - 9.08x10^-3 M
pH = -log [9.165x10^-3 - 9.08x10^-3] =1.74
Answer: New pH will be 4.06 .
Explanation: The problem could easily be solved using Handerson equation. Handerson equaton is used to calculate the pH of buffer solution.
[tex]pH=pK_a+log(\frac{base}{acid})[/tex]
For the given problem, the buffer solution is a mixture of a weak acid(benzoic acid) and a salt of it, known as conjugate base(sodium benzoate).
When a strong acid is added to the buffer solution then it reacts with the base(benzoate ion) present in the buffer solution and produce a the weak acid(benzoic acid).
The reaction could be shown as:
[tex]C_6H_5COO^-(aq)+H^+(aq)\rightleftharpoons C_6H_5COOH(aq)[/tex]
First of all we calculate the initial moles of acid and base originally present in the buffer solution and for this the volume is multiplied by the molarity.
initial moles of benzoic acid = [tex]0.250L(\frac{0.250mol}{1L})[/tex]
= 0.0625 moles
initial moles of benzoate ion = [tex]0.250L(\frac{0.20mol}{1L})[/tex]
= 0.0500 moles
moles of HCl or [tex]H^+[/tex] added to the buffer = [tex]25.0mL(\frac{1L}{1000mL})(\frac{0.100mol}{1L})[/tex]
= 0.0025 moles
From the equation we have written above, HCl and benzoate ion react in 1:1 mol ratio. As HCl is limiting reactant, 0.0025 moles of it will react with exactly 0.0025 moles of benzoate ion and form 0.0025 moles of benzoic acid.
So, the moles of benzoic acid after addtion of HCl = 0.0625 + 0.0025 = 0.065 moles
moles of benzoate ion after addition of HCl = 0.0500 - 0.0025 = 0.0475 moles
Total volume of the solution = 0.250 L + 0.025 L = 0.275 L
concentration of benzoic acid = [tex]\frac{0.065mol}{0.275L}[/tex]
= 0.236M
concentration of benzoate ion = [tex]\frac{0.0475mol}{0.275L}[/tex]
= 0.173M
Pka is calcuulated from the given Ka value as:
[tex]pK_a=-logK_a[/tex]
[tex]pk_a=-log(6.5*10^-^5)[/tex]
[tex]pK_a[/tex] = 4.19
Let's plug in the values in the Handerson equation:
[tex]pH=4.19+log(\frac{0.173}{0.236})[/tex]
pH = 4.19 - 0.13
pH = 4.06
So, the pH of the solution after an addition of HCl will be 4.06 .
