Question

A 0.0720 L volume of 0.132 M hydrobromic acid (HBr), a strong acid, is titrated with 0.264 M potassium hydroxide (KOH), a strong base. Determine the pH at the following points in the titration: (a) before any KOH has been added. (b) after 0.0180 L KOH has been added. (c) after 0.0360 L KOH has been added. (d) after 0.0540 L KOH has been added.

Answer 1

Because it is a strong acid-base reaction, the reaction will be:

HBr + NaOH ===> H2O + NaCl

Initial concentration of HBr = 0.072 x 0.132 = 0.009504 Moles H+

pH = -log H+

pH = -log 0.009504

pH = 2.02

(b) after 0.0180 L KOH has been added

No of Moles of KOH is added = 0.018 x 0.264 = 0.004752 Moles

Which results in: 0.009504 - 0.004752 = 0.004752 Moles

H+ = 0.004752

pH = 2.32

(c) after 0.0360 L KOH has been added.

No of Moles of KOH is added = 0.036x 0.264 = 0.009504 Moles

Which results in: 0.009504 - 0.009504 = 0 Moles

all acids get Neutrlized

pH = 7

after 0.0540 L KOH has been added.

No of Moles of KOH is added = 0.054x 0.264 = 0.014256 Moles

Which results in: 0.009504 - 0.009504  = 0.4752 OH- Moles

pOH = -Log OH-

pOH = 2.32

pH = 14-2.32 = 11.67