Asked by stephanie

1L of a buffer composed of acetic acid and sodium acetate has a pH of 4.3. Adding 10mL of 2M sodium hydroxide solution to 100mL of this buffer causes the pH to rise to 4.87. what is the total molarity of the original buffer?
Answered by DrBob222
pH = pka + log b/a
4.3 = 4.74 + log b/a
b/a = 0.363
----------------
add 100 mL x 2M = 20 mmols to obtain
4.87 = 4.74 + log b/a
b/a = 1.35 and redo to read
1.349 = (b+20)/(a-20)
1.349(a-20) = b+20
1.349a - 27 = b+20
Substitute 0.363 for b to obtain
1.349a - 27 = 0.363a + 20
0.986a = 47
a = 47.7
b = 0.363a = 0.363*47.7 = 17.3
a + b = 47.7 + 17.3 = 65
--------------------
Check:
..........HAc + OH^- ==> Ac^- + H2O
I.........47.7...0.......17.3
add ............20.........
C.........-20..-20.......+20
E.........27.7...0.......37.3

pH = 4.74 + log(37.3/27.7)
pH = 4.87

Answered by Ashley
i got a total concentration of 0.0774M is this correct?
Answered by Anonymous
Where are you getting that?
There are no AI answers yet. The ability to request AI answers is coming soon!