1) If 25.2 mL of 0.109 M acid with a pKa of 5.55 is titrated with 0.102 M NaOH s
ID: 560888 • Letter: 1
Question
1) If 25.2 mL of 0.109 M acid with a pKa of 5.55 is titrated with 0.102 M NaOH solution, what is the pH of the titration mixture after 13.7 mL of base solution is added?2) If 28.8 mL of 0.108 M acid with a pKa 4.15 is titrated with 0.108 M NaOH solution, what is the pH of the acid solution before any base solution is added?
3)If 27.9 mL of 0.107 M acid with a pKa of 5.48 is titrated with 0.106 M NaOH solution, what is the pH of the titration mixture at halfway to the equivalence point? 1) If 25.2 mL of 0.109 M acid with a pKa of 5.55 is titrated with 0.102 M NaOH solution, what is the pH of the titration mixture after 13.7 mL of base solution is added?
2) If 28.8 mL of 0.108 M acid with a pKa 4.15 is titrated with 0.108 M NaOH solution, what is the pH of the acid solution before any base solution is added?
3)If 27.9 mL of 0.107 M acid with a pKa of 5.48 is titrated with 0.106 M NaOH solution, what is the pH of the titration mixture at halfway to the equivalence point? 1) If 25.2 mL of 0.109 M acid with a pKa of 5.55 is titrated with 0.102 M NaOH solution, what is the pH of the titration mixture after 13.7 mL of base solution is added?
2) If 28.8 mL of 0.108 M acid with a pKa 4.15 is titrated with 0.108 M NaOH solution, what is the pH of the acid solution before any base solution is added?
3)If 27.9 mL of 0.107 M acid with a pKa of 5.48 is titrated with 0.106 M NaOH solution, what is the pH of the titration mixture at halfway to the equivalence point?
Explanation / Answer
Q1)
pKa of acid = 5.55
HA + NaOH ------------> NaA + H2O
25.2x0.109=2.7468 13.7x0.102=1.3974 0 0 initial mmoles
1.3494 0 1.3974 - after reaction
So the solution forms a buffer of weak acid and its conjugate base.The pH of buffer is given by Hendersen equation as
pH = pKa + log [conjugate base]/[acid]
= 5.55 + log 1.3974/1.3494
= 5.565
Q2) molarity of acid = 0.108M
pKa of acid = 4.15
pH of the acid before any base is added is given by
pH = 1/2[pKa -logC]
= 1/2[ 4.15 -log 0.108]
= 5.1165
Q3) At the equivalence
HA + NaOH ------------> NaA + H2O
27.9x 0.107= 2.985 0 0 - initial mmoles
1.492 0 1.492 - at half equivalence
thus the solution is a buffer
pH = Pka + log[conjugatebase]/[acid]
= 5.48 + log 1.492/1.492
= 5.48
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.