Question

For all of the following questions 20.00 mL of 0.192 M HBr is titrated with 0.200 M KOH.
Region 1: Initial pH: Before any titrant is added to our starting material
What is the concentration of H+ at this point in the titration?
M

What is the pH based on this H+ ion concentration?

Region 2: Before the Equivalence Point 10.13 mL of the 0.200 M KOH has been added to the starting material.
Complete the BCA table below at this point in the titration. (Be sure to use moles)
HBr (aq)   KOH (aq)   →   H2O (l)   KBr (aq)
B               NA  
C               NA  
A               NA  
From the moles of HBr left after the reaction with KOH what will the pH be at this point in the titration?

Region 3: Equivalence Point
What volume of the titrant has been added to the starting material at the equivalence point for this titration?
mL

At the equivalence point an equal number of moles of OH- and H+ have reacted, producing a solution of water and salt. What affects the pH at this point for a strong-acid/strong-base titration?
   The basicity of the salt anion
   The acidity of the salt cation
   None of these
   The auto-ionization of water

Region 4: After the Equivalence Point 30.99 mL of the 0.200 M KOH has been added to the starting material
Complete the BCA table below at this point in the titration. (Use moles)
HBr (aq)   KOH (aq)   →   H2O (l)   KBr (aq)
B               NA  
C               NA  
A               NA  
From the moles of KOH remaining after the reaction with HBr what is the pOH at this point in the titration?

Calculate the pH of the solution from the pOH found in the previous step

Answer 1

millimoles of HBr = 20 x 0.192 = 3.84

a) before addition of any KOH

[HBr] = 0.192 M

[H+] = 0.192 M

pH = -log [H+] = -log (0.192)

pH = 0.72

b) region 2

millimoles of KOH = 0.20 x 10.13 = 2.03

here millimoles of HBr > millimoles of KOH

[H+] = 3.84 - 2.03 / 10.13 + 20 = 0.060 M

pH = -log (0.060)

pH = 1.22

region 3 :

this is equivalence point .

At equivalence point ,

pH = 7.00.

volume of titrant added = 19.2 mL

region 4 )

millimoles of KOH = 6.20

millimoles of KOH > millimoles of HBr

[OH-] = 6.20 - 3.84 / (20 + 30.99) = 0.0462

pOH = -log (0.0462)

pOH = 1.33

pH = 12.66