a buffer solution that is 0.100 M acetate ion and 0.100 M acetic is prepared.
calculate the initial pH, final pH, and chanage in pH when 4.0 ml of 0.20 M NaOH is added to 100.0 mL of the buffer
init pH=
final pH=
change pH=
calculate the initial pH, final pH, and chanage in pH when 4.0 ml of 0.20 M NaOH is added to 100.0 mL of the buffer
init pH=
final pH=
change pH=
-
According to Henderson-Hasselbalch equation: pH = pKa + log [A-]/[HA]
So initial pH = 4.75 + log (0.10/0.1). Since log1 = 0, then initial pH = pKa = 4.75
Number of moles of NaOH added = 0.004L x 0.20M = 8 x 10^-4M. The added moles NaOH would neutralize the same number of moles of the acid hence the new concentration of the acid would be 0.10 - 0.0008 = 0.0992M. Hence the final pH = 4.75 + log(0.0992/0.10) = 4.75 + 0.0035 = 4.7535 ~ 4.754.
Change in pH = 4.754 - 4.75 = 0.004
So initial pH = 4.75 + log (0.10/0.1). Since log1 = 0, then initial pH = pKa = 4.75
Number of moles of NaOH added = 0.004L x 0.20M = 8 x 10^-4M. The added moles NaOH would neutralize the same number of moles of the acid hence the new concentration of the acid would be 0.10 - 0.0008 = 0.0992M. Hence the final pH = 4.75 + log(0.0992/0.10) = 4.75 + 0.0035 = 4.7535 ~ 4.754.
Change in pH = 4.754 - 4.75 = 0.004