Question

Consider the titration of 100.0 mL of 0.100 M NaOH with 1.00 M HBr. Find the pH at the following volumes of acid added.

Va = 0 mL
Va = 1.0 mL
Va = 5.0 mL
Va = 9.0 mL
Va = 9.9 mL
Va = 10.0 mL
Va = 10.1 mL
Va = 12.0 mL

Make a graph of pH versus V_a = 0, 1.0, 5.0, 9.0, 9.9, 10.0, 10.1, and 12.0 mL.

Answer 1

millimoles of NaOH = 100 x 0.1 = 10

a) before addition of any HBr

NaOH = 0.1 M

[OH-] = 0.1 M

pOH = -log[OH-] = -log (0.1) = 1

pH +pOH = 14

pH = 13

b) 1.00 ml HBr added

millimoles of HBr = 1 x 1 = 1

millimoles of base > millimoles of acid

so [OH-] = (base millimoles - acid millimoles )/ total volume

               = (10 - 1 ) / (100 + 1)

              = 0.0891 M

pOH = 1.05

pH = 12.95

c) 5.0 ml HBr added

millimoles of HBr = 5 x 1 = 5

millimoles of base > millimoles of acid

so [OH-] = (base millimoles - acid millimoles )/ total volume

               = (10-5) / (100 + 5)

              = 0.0476 M

pOH = 1.32

pH = 12.68

d) 9 ml HBr added

millimoles of HBr = 9 x 1 = 9

millimoles of base > millimoles of acid

so [OH-] = (base millimoles - acid millimoles )/ total volume

               = (10 - 9) / (100 + 9)

              = 0.00917 M

pOH = 2.04

pH = 11.96

e) 9.9 ml HBr added

millimoles of HBr = 9.9 x 1 = 9.9

millimoles of base > millimoles of acid

so [OH-] = (base millimoles - acid millimoles )/ total volume

               = (10 - 9.9) / (100 + 9.9)

              = 9.099 x 10^-4 M

pOH = 3.04

pH = 10.96

f) 10 ml HBr added

millimoles of base = millimoles of acid

so it equivalence point . strong acid + strong base

pH = 7.00

e) 10.1 ml HBr added

millimoles of HBr = 10.1 x 1= 10.1

millimoles of base < millimoles of acid

so [H+] = ( acid millimoles -base millimoles)/ total volume

               = (10.1 - 10 ) / ( 100 + 10.1)

              = 9.08 x 10^-4 M

pH = -log[H+]

pH = 3.04

f) 12 ml HBr added

millimoles of HBr = 12 x 1= 12

millimoles of base < millimoles of acid

so [H+] = ( acid millimoles -base millimoles)/ total volume

               = (12 - 10) / ( 100 + 12)

              = 0.01786 M

pH = -log[H+]

pH = 1.75