Question

Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.170 M NaOH(aq), with 0.170 M HI(aq). _______ Number Note: Enter your answers with two decimal places. (a) before addition of any HI Number (b) after addition of 13.5 mL of HI Number (c) after addition of 23.5 mL of HI Number (d) after the addition of 35.0 mL of HI Number (e) after the addition of 41.5 mL of HI Number (f) after the addition of 50.0 mL of HI

Answer 1

millimoles of NaOH = 0.170 x 35 = 5.95

a) before addition of any HI

NaOH = 0.170 M

[OH-] = 0.170 M

pOH = -log[OH-] = -log (0.170) = 0.77

pH +pOH = 14

pH = 13.23

b) 13.5 ml HI added

millimoles of HI = 13.5 x 0.170 = 2.295

millimoles of base > millimoles of acid

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

               = (5.95 -2.295) / (35+13.5)

              = 0.0753M

pOH = 1.12

pH = 12.88

c) 23.5 ml HI added

millimoles of HI = 23.5 x 0.170 = 3.995

millimoles of base > millimoles of acid

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

               = (5.95 -3.995) / (35+23.5)

              = 0.033M

pOH = 1.48

pH = 12.52

d) 35 ml HI added

millimoles of HI = 35 x 0.170 = 5.95

millimoles of base = millimoles of acid

so it equivalence point . strong acid + strong base

pH = 7.00

e) 41.5 ml HI added

millimoles of HI = 41.5 x 0.170 = 7.055

millimoles of base < millimoles of acid

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

               = (7.055 - 5.95 ) / (35+41.5)

              = 0.0144M

pH = -log[H+]

pH = 1.84

f) 50 ml HI added

millimoles of HI = 50 x 0.170 = 8.5

millimoles of base < millimoles of acid

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

               = (8.5 - 5.95 ) / (35+50)

              = 0.03M

pH = -log[H+]

pH = 1.52