Question

Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make 1.00 L of solution. Buffer A is prepared using 1.00 mol each of formic acid and sodium formate. Buffer B is prepared by using 0.010 mol of each. (Ka(HCOOH)=1.8×10−4.)

Calculate the change in pH for each buffer upon the addition of 10 mL of 1.00M HCl.

Answer 1

Buffer A

pKa = -logKa = 3.75

pH = pKa + log(conjugate/acid)

pH = 3.75+ log(1/1)

Buffer B

pKa = -logKa = 3.75

pH = pKa + log(conjugate/acid)

pH = 3.75+ log(0.01/0.01)

after addition of

mmol of HCl = MV = 10*1 = 10 mmol or = 0.01 mol

then

for Buffer A)

recall that acid concnetation increases as conjugate decreases so

acid = 1+0.01 = 1.01

conjugate = 1-0.01 = 0.99

then

pH = 3.75+ log(1.01/0.99) = 3.7586

ph = 3.7586

for Buffer B)

recall that acid concnetation increases as conjugate decreases so

acid = 0.01+0.01 = 1.01

conjugate = 0.01-0.01 = 0

meaning that the buffer is "destroyed"

there is only acid in solutoin so

[H+] = mmol of H+ / total V

Total V = 1L + 10 ml = 1.01

mmol of H+ = 10 mmol = 0.001 mol

[H+] = 0.001/1.01

then

pH =-log(0.000990099) = 3.00432