For all of the following questions 20.00 mL of 0.195 M HBr is titrated with 0.20

Question

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

Explanation / Answer

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

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