Question

Calculate the hypothetical pH AFTER addition of 40.00 mL of reagent for the titration of 50.00 mL of 0.0800 M HCl with 0.1000 M NaOH at 25°C. Auranmaa

Answer 1

1)

Given:

M(HCl) = 0.08 M

V(HCl) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 40 mL

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

mol(HCl) = 0.08 M * 50 mL = 4 mmol

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

mol(NaOH) = 0.1 M * 40 mL = 4 mmol

We have:

mol(HCl) = 4 mmol

mol(NaOH) = 4 mmol

4 mmol of both will react to form neutral solution

hence pH of solution will be 7

Answer: 7.00

2)

[OH-]= 0.0800 M

use:

pOH = -log [OH-]

= -log (0.0800)

= 1.0969

use:

PH = 14 - pOH

= 14 - 1.0969

= 12.9031

Answer: 12.90