First determine pKa.
pH = pKa + log(A^-)/(HA)
I have approximately 3 but you need to be more accurate than that.
650 mL x 0.271M = about 170 mmoles
650 mL x 0.150M = about 95 mmoles
mmoles LiOH added = 130 x 0.155M = about 20 mmoles.
...........HA + OH^- ==> A^- + H2O
initial...170....0........95
added...........20
change...-20....-20......+20
equil....150.....0........115
Substitute the equil values (after refining them) into the HH equation and solve for pH.
A buffer that is 0.271 M in acid, HA, and 0.150 M in the potassium salt of its conjugate base, KA, has a pH of 2.85. What is the pH of the buffer after 130. mL of 0.155 M LiOH is added to 0.650 L of this buffer? Assume that the volumes are additive.
1 answer