a) when adding to 24 mL of 0.19 M NaOH:
first, we need to get moles of HCl = molarity * volume
= 0.19 M * 0.019 L
= 0.00361 moles
then, moles of NaOH = molarity * volume
= 0.19 M * 0.024 L
= 0.00456 moles
NaOH remaining = 0.00456 - 0.00361
= 0.00095 moles
when the total volume = 24mL+19 mL = 43 mL
∴ molarity of NaOH = moles / total volume
= 0.00095 / 0.043L
= 0.0221 M
when POH = -㏒[OH-]
= -㏒0.0221 M
= 1.66
∴PH = 14 - POH
= 14- 1.66
= 12.34
b) when adding to 29 mL of 0.24 M NaOH:
when moles of HCl = 0.00361 moles
then, moles of NaOH = molarity * volume
= 0.24 M * 0.029 L
= 0.00696 moles
∴ NaOH remaining = 0.00696 - 0.00361
= 0.00335 moles
the total volume = 29 mL + 19 mL = 48 mL
molarity of NaOH = moles / total volume
= 0.00335 / 0.048L
= 0.0698 M
∴POH = -㏒[OH-]
= -㏒ 0.0698
= 1.16
∴ PH = 14- POH
= 14- 1.16
= 12.84
