Question
100 mL of buffer solution that is 0.15M HA (Ka= 6.8x10-5) and 0.20M NaA is mixed with 21mL of 0.35M NaOH. What is the pH of the resulting solution?
Answers
Anonymous
NaOH will react with HA.
Calculate moles, then:
0.00735 mole of NaOH+ 0.015 NaA = x : new amount of NaA
0.020 HA-0.00735 moles of NaOH=y: new amount of HA
-log(KA)=pka
pH=pka+log[x/y]
Calculate moles, then:
0.00735 mole of NaOH+ 0.015 NaA = x : new amount of NaA
0.020 HA-0.00735 moles of NaOH=y: new amount of HA
-log(KA)=pka
pH=pka+log[x/y]