chemistry
posted by stephanie on .
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?

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)/(a20)
1.349(a20) = 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 
i got a total concentration of 0.0774M is this correct?

Where are you getting that?