Question

In lab you are asked to titrate 50.00 mL of 1.000 M HCl with 1.000 M NaOH. Predict what the pH will be after adding 60.00 mL of NaOH.

Answer 1

Given:

M(HCl) = 1 M

V(HCl) = 50 mL

M(NaOH) = 1 M

V(NaOH) = 60 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 1 M * 50 mL = 50 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 1 M * 60 mL = 60 mmol

We have:

mol(HCl) = 50 mmol

mol(NaOH) = 60 mmol

50 mmol of both will react

remaining mol of NaOH = 10 mmol

Total volume = 110.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 10 mmol/110.0 mL

= 9.091*10^-2 M

use:

pOH = -log [OH-]

= -log (9.091*10^-2)

= 1.0414

use:

PH = 14 - pOH

= 14 - 1.0414

= 12.9586

Answer: 12.96