Question

40ml of a 2.00M HBr solution was added to 100mL of a 1.0 M LiOH solution. what is the pH if the combined solution?

Answer 1

Given:

M(HBr) = 2 M

V(HBr) = 40 mL

M(LiOH) = 1 M

V(LiOH) = 100 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 2 M * 40 mL = 80 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 1 M * 100 mL = 100 mmol

We have:

mol(HBr) = 80 mmol

mol(LiOH) = 1*10^2 mmol

80 mmol of both will react

remaining mol of LiOH = 20 mmol

Total volume = 140.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 20 mmol/140.0 mL

= 0.1429 M

use:

pOH = -log [OH-]

= -log (0.1429)

= 0.8451

use:

PH = 14 - pOH

= 14 - 0.8451

= 13.1549

Answer: 13.15