what is the pH of the buffer formed by adding 100 mL of 0.150 M potassium hydrog
ID: 735668 • Letter: W
Question
what is the pH of the buffer formed by adding 100 mL of 0.150 M potassium hydrogen phthalate to(a) 100 mL of 0.0800 M NaOH
(b) 100 mL of 0.0800 M HCl
Explanation / Answer
a) Near enough, all the added NaOH is used up in the reaction HP- + OH- --> P2- + H2O (where H2P is phthalic acid) You are left with 0.00800 mol P2- and 0.0070 mol unreacted KHP in 200 mL Use the Henderson-Hasselbach equation, or the definition of Ka, to find pH. The relevant Ka is that for HP- as an acid, i.e. 2nd ka of H2P. b) Interesting Since HCl is a strong acid, you will have the reaction HP- + H+(aq) (from HCl) --> H2P going virtually to completion. So you now have 0.00800 mmol H2P, and 0.0070 mol HP- H-H Equation again, but this time considering the FIRST dissociation of H2P The trick here is to remember that HP- can be considered either as a weak acid, or as a weak base.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.