NaOH will react with HA.
Calculate moles, then:
0.00532 mole of NaOH+ 0.020 NaA = x : new amount of NaA
0.0150 HA-0.00532 moles of NaOH=y: new amount of HA
-log(KA)=pka
pH=pka+log[x/y]
Second problem is essentially the same:
HCl is added to the amount of HA and subtracted from the amount of NaA.
100 mL of buffer solution that is 0.15M HA (Ka= 6.8x10-5) and 0.20M NaA is mixed with 15.2mL of 0.35M NaOH. What is the pH of the resulting solution?
100 mL of buffer solution that is 0.15M HA (Ka= 6.8x10-5) and 0.20M NaA is mixed with 28.3mL of 0.25M HCl. What is the pH of the resulting solution?
1 answer
1 answer