base = b
acid = a
First you must determine the (acid) and (base). You do that by solving two equations simultaneously.
b+a = 0.1M is equation 1
pH = pKa + log (acid)/(base) and substitute like this from the problem
5.00 = 4.76 + log (b/a) is equation 2.
Solve the two for a and b. These aren't the exact numbers but they're close.
a = about 0.04
b = about 0.06
millimols acid = 0.04*140 = about 5.6
mmols base = 0.06* 140 = 8.4
mmols HCl added = about 2.6
........b + H^+ ==> Hb
I......8.4..0.......5.6
add........2.6................
C.....-2.6.-2.6......+2.6
E.....5.8...0.......8.2
Substitute the E line into the HH equation and solve for pH.
Post your work if you get stuck.
Check this work carefully. I did it in a hurry.
A beaker with 1.40×102mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 7.10mL of a 0.370M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
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