A buffer contains 0.16mol of propionic acid (C2H5COOH) and 0.17mol of sodium pro
ID: 793945 • Letter: A
Question
A buffer contains 0.16mol of propionic acid (C2H5COOH) and 0.17mol of sodium propionate(C2H5COONa) in 1.20 L.
1)What is the pH of this buffer?
2)What is the pH of the buffer after the addition of 0.02mol of NaOH?
3)What is the pH of the buffer after the addition of 0.02mol of HI?
A sample of 7.80L of NH3 (ammonia) gas at 22 ?Cand 735 torr is bubbled into a 0.350L solution of 0.400 M HCl (hydrochloric acid).
The Kb value for NH3 is 1.8
A buffer contains 0.16mol of propionic acid (C2H5COOH) and 0.17mol of sodium propionate(C2H5COONa) in 1.20 L. What is the pH of this buffer? What is the pH of the buffer after the addition of 0.02mol of NaOH? What is the pH of the buffer after the addition of 0.02mol of HI? A sample of 7.80L of NH3 (ammonia) gas at 22 ?Cand 735 torr is bubbled into a 0.350L solution of 0.400 M HCl (hydrochloric acid). The Kb value for NH3 is 1.8 times 10?5. Assuming all the NH3 dissolves and that the volume of the solution remains at 0.350L, calculate the pH of the resulting solution.Explanation / Answer
A buffer solution may be acidic or basic. Your question involves the acidic buffer. An acidic buffer solution contains a weak acid and its salt.
In your question the weak acid is propionic acid;
HC3H5O2 (or C2H5COOH) .... Ka = 1.3x10^-5
and its salt is sodium propionate (NaC3H5O2) (or C2H5COONa).
As these formulas imply propionic acid is a monoprotic acid.
Concentrations of the substances;
[HC3H5O2] = 0.16 mol / 1.20 L = 0.133 M
[NaC3H5O2] = 0.17 mol / 1.20 L = 0.142 M
(a) Dissociations:
HC3H5O2(aq) <-----> H+(aq) + C3H5O2-(aq)
0.133 - x M ................ x M ............ x M
NaC3H5O2(aq) -----> Na+(aq) + C3H5O2-(aq)
0.142 M ....................0.142 M .....0.142 M
C3H5O2- is the common ion, but the concentration of C3H5O2- ion produced by the dissociation of the salt (NaC3H5O2) will be very large compared to the concentration of C3H5O2- ion produced by the dissociation of the acid, and hence x is neglected.
Ka = [H+][C3H5O2-] / [HC3H5O2]
This expression in some cases simplified as (since x is neglected)
Ka = [H+][SALT] / [ACID]
Substituting the values;
1.3x10^-5 = [H+] (0.142) / (0.133)
[H+] = (1.3x10^-5 x 0.133) / 0.142 = 1.22x10^-5
pH = -log[H+] = -log[1.22x10^-5] = 4.91
(b) When 0.02 mol NaOH is added, 0.02 mol acid is neutralized and 0.16 - 0.02 = 0.14 mol acid is left.
HC3H5O2(aq) + NaOH(aq) ----> NaC3H5O2(aq) + H2O(l)
New mole of acid: 0.16 - 0.02 = 0.14 mol
New molarity of acid : 0.14 mol / 1.20 L = 0.117 M
New mole of salt : 0.17 + 0.042 = 0.19 mol
New molarity of salt : 0.19 mol / 1.20 L = 0.158 M
Ka = [H+][SALT] / [ACID]
[H+] = Ka x [ACID] / [SALT]
[H+] = (1.3x10^-5 x 0.117) / 0.158 = 9.6x10^-6
pH = 5.018
(c) In our original buffer solution;
[H+] = 1.22x10^-5 M
mole of H+ from the dissociation of added HI acid = 0.02 mol
Molarity of added H+ ion = 0.02 mol / 1.20 L = 0.0167 M
HC3H5O2(aq) <-----> H+(aq) + C3H5O2-(aq)
The added H+ ions will shift the equilibrium left.
Reaction : HC3H5O2(aq) <-----> H+(aq) + C3H5O2-(aq)
initial: ........ 0.133 M ................. 1.43x10^-5 M .... 0.142 M
added : ........................................ 0.0095 M
change : .. +0.0167 M .............. - 0.0095 M ... - 0.0167 M
new equil: 0.1497 M ................... ????? .......... 0.1253 M
[H+] = Ka x [ACID] / [SALT]
[H+] = (1.3x10^-5 x0.1497) / 0.1253= 0.024x10^-5
pH =6.62
The main characteristics of buffer solutions is illustrated in this question:
When a strong acid or a base is added to a buffer solution, its pH does not change considerably.
Original buffer, pH = 4.91
When strong base is added, pH = 5.018
When strong acid is added, pH = 6.62
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