Consider a buffer solution that contains 0.25 M C6H4(CO2H)(CO2K) and 0.15 M C6H4
ID: 1069363 • Letter: C
Question
Consider a buffer solution that contains 0.25 M C6H4(CO2H)(CO2K) and 0.15 M C6H4(CO2K)2. pKa(C6H4(CO2H)CO2-)=5.41.
a) Calculate the change in pH if 0.140 g of solid NaOH is added to 190 mL of this solution.a
b) If the acceptable buffer range of the solution is ±0.10 pH units, calculate how many moles of H3O+ can be neutralized by 250 mL of the initial buffer.
Consider a buffer solution that contains 0.25 M C6H4(CO2H)(CO2K) and 0.15 M C6H4(CO2K)2. pKa(C6H4(CO2H)CO2 5.41 Correct 1. Calculate its pH 5.19Explanation / Answer
First of All pH of buffer solution that contains 0.25 M C6H4(CO2H)(CO2K) and 0.15 M C6H4(CO2K)2. where pKa of (C6H4(CO2H)CO2-)=5.41.
Now pH of Solution will be
pH = pKa + log(A-/HA)
pKa = 5.41
pH = 5.41 + log(0.15 / 0.25)
pH = 5.41 - 0.2218
pH = 5.19
Now pH if 0.140 g of solid NaOH is added to 190 mL of this solution means 0.0035 mol(0.140/40) of solid NaoH is added to 0.0475 mol (0.25 x 0.190) buffer soution
so
The equilibrium of the solution can be shown in a shorthand notation:
C6H4(CO2H)(CO2K)(aq) == H+ (aq) + C6H4(2CO2- )(aq)
Initial: 0.0044 0 0.051
Change: -x x x
Equilibrium: 0.0044 - x x 0.051 + x
Now pKa= 5.41
5.41 = -log (Ka)
Ka = 4.5 x 10^-3
Ka = 4.5 x 10^-3 = [H+][ C6H4(2CO2- )] / [C6H4(CO2H)(CO2K)(aq)] = (x) (0.051 + x) / (0.0044 - x) =(x)(0.051) / 0.0044
Thus
11.59 x = 4.5 x 10^-3
x = 38 x 10^-5
pH = - log (38 x 10^-5)
pH = 3.42
The change in pH from adding 0.0035 mol OH- to the buffered solution is only
5.19 - 3.42 = 1.77
If 0.0035 mol solid NaOH is added to 1.0L water, the pH will change from 7.00 to 12.00 (+5.00).
[H+] = x = 1.7 x 10-5 M
pH = 4.76
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