100 mL of buffer solution that is 0.15M HA (Ka= 6.8x10-5) and 0.20M NaA is mixed...
100 mL of buffer solution that is 0.15M HA (Ka= 6.8x10-5) and 0.20M NaA is mixed with 18.2mL of 0.35M NaOH. What is the pH of the resulting solution?
Homework Answers
pKa of acid = -log Ka = -log 6.8x10-5 =4.1675
The buffer is
HA + NaOH ----------------------> NaA + H2O
100x0.15=15 0 100x 0.20=20 - ---- initial mmoles
--------- 18.2 x0.35=6.37 ---------- ------- change
15-6.37=8.63 0 20+6.37=26.37 ------- equilibrium mmoles
Thus the solution is still a buffer and its pH is calculated using Hendersen equation as
pH = pKa + log [conjugate base]/[acid]
=4.1675 + log 26.37/8.63 =4.6526
The pH of the resulting solution = 4.6526
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