Question

Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and...

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.

Homework Answers

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

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