In: Chemistry
For all of the following questions 20.00 mL of 0.195 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 5.68 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 acidity of the salt cation
The basicity of the salt anion
The auto-ionization of water
None of these
Region 4: After the Equivalence Point 31.31 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
HBr + KOH --> NaBr + H2O
Region 1, [H+] = 0.195 M
pH = -log(0.195) = 0.71
Region 2, KOH added = 0.2 M x 5.68 ml = 1.14 mmol
moles HBr (initial) = 0.195 M x 20 ml = 3.90 mmol
HBr + KOH --> H2O + KBr
I 3.90 - - -
C -1.14 - 1.14 1.14
E 2.76 - 1.14 1.14
pH = -log(2.76/25.68 ml) = 0.97
Region 3, Equivalence
Volume KOH added = 3.90 mmol/0.20 M = 19.5 ml
pH = 7
The pH is dependent upon,
The auto-ionization of H2O
Region 4, 31.31 ml KOh added
excess [OH-] = (31.31 - 19.5) ml x 0.2 M/51.31 ml = 0.046 M
[H+] = 1 x 10^-14/0.046 = 2.2 x 10^-13 M
pH = -log(2.2 x 10^-13) = 12.66