A particular acid [HA-] with pKa=5.64 is used to make a buffered solution with a pH of 5.28 ; the total salt concentration is 43 mM. U?

Question

A particular acid [HA-] with pKa=5.64 is used to make a buffered solution with a pH of 5.28 ; the total salt concentration is 43 mM. How many millimoles of conjugate base [A-] were used? Hint: Use the Henderson-Hasselbalch equation and this equation [HA] + [A-] = 43

Answer 1

pH = pKa + log [A-] / [HA]

5.28 = 5.64 + log [A-] / [HA] 

log [A-] / [HA] = -0.36

taking antilogs

[A-] / [HA] = 10^- 0.36 =0.436515832    ,,, eq 1 

[HA] + [A-] = 43 mM.....given 

substituting in 1 

[A-] / 43 - [A-] = 0.436515832

[A-] = 0.436515832 (43 - [A-])

[A-] = 18.77018079 - 0.436515832 [A-] 

1.436515832 [A-] = 18.77018079

[A-] = 13.07 mM

[HA] = 29.93 mM

testing 

5.28 = 5.64+ log( 0.436515832 ) = 5.28 

your answer [A-] = 13.07 mM