Calculate the ph of the resulting solution if 19.0 ml of 0.190 m hcl (aq) is added to

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