A 45.0 mL buffer solution is 0.430 M Na2HPO4 (aq) and 0.330 M KH2PO4 (aq). The k
ID: 712952 • Letter: A
Question
A 45.0 mL buffer solution is 0.430 M Na2HPO4 (aq) and 0.330 M KH2PO4 (aq). The ka for H2PO4- is 6.2*10-8
a. what is the pH of the buffer solution
b. what are the pH and the pH change resulting from the addition of 9.0*10-4 mol of NaOH to the buffer solution
c. what are the pH and the pH change resulting from the addition of 9.5*10-4 mol of nitric acid to the initial buffer system
d. how many mL of 0.400 M sodium hydroxide would you have to add to the buffer in order to exhaust its buffering capacity
Explanation / Answer
a)
mmoles of H2PO4- = 0.330 x 45 = 14.85
mmoles of HPO42- = 0.430 x 45 = 19.35
pKa = 7.21
pH = pKa + log [salt / acid]
= 7.21 + log [19.35 / 14.85]
pH = 7.32
b)
mmoles of H2PO4- = 0.330 x 45 = 14.85
mmoles of HPO42- = 0.430 x 45 = 19.35
mmoles of NaOH = 0.9
pH = pKa + log [salt + C / acid - C]
= 7.21 + log [19.35 + 0.9 / 14.85 - 0.9]
pH = 7.37
pH change = 7.37 - 7.32 = 0.05
pH change = 0.05
c)
mmoles of HNO3 = 0.95
pH = pKa + log [salt - C / acid + C]
= 7.21 + log [19.35 - 0.95 / 14.85 + 0.95]
pH = 7.28
pH change = - 0.04
d)
mmoles of H2PO4- = 14.85
mmoles of H2PO4 = mmoles of NaOH . then buffer fails.
14.85 = 0.4 x V
V = 37.125
volume of NaOH = 37.12 mL
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.