HA(aq)+OH(aq)A(aq)+H2O(l) A certain weak acid, HA, with a K a value of 5.61×106,

Question

HA(aq)+OH(aq)A(aq)+H2O(l)

A certain weak acid, HA, with a Ka value of 5.61×106, is titrated with NaOH.

Part A

A solution is made by titrating 9.00 mmol (millimoles) of HA and 2.00 mmol of the strong base. What is the resulting pH?

Express the pH numerically to two decimal places.

Part B

More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 43.0 mL ?

Express the pH numerically to two decimal places.

A titration involves adding a reactant of known quantity to a solution of an another reactant while monitoring the equilibrium concentrations. This allows one to determine the concentration of the second reactant. The equation for the reaction of a generic weak acid HA with a strong base is

HA(aq)+OH(aq)A(aq)+H2O(l)

A certain weak acid, HA, with a Ka value of 5.61×106, is titrated with NaOH.

Part A

A solution is made by titrating 9.00 mmol (millimoles) of HA and 2.00 mmol of the strong base. What is the resulting pH?

Express the pH numerically to two decimal places.

pH = ______

Part B

More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 43.0 mL ?

Express the pH numerically to two decimal places.

pH = ______

Explanation / Answer

HA + OH- = A- + H2O
mmol HA = 9.00 - 2.00 = 7.00

mmol A- = 2.00
pKa = - log Ka

Ka = 5.61×106

So

Pka = -log Ka = -log (5.61×106)= 5.25

According to Henderson –hasselbach equation

p H = p K a + l o g ( [ A ] [ H A ]

pH = 5.25 + log 2.00/7.00=4.71

Part B

1mmol = 0.001 mol

Volume = 43 ml
[A-] = 0.0090 mol/ 0.0430 L=0.209 M
A- + H2O <-----> HA + OH-
Kb = Kw/Ka = 1.00 x 1014/ 5.61×106 = 1.78 x 10 -10

1.78 x 10-10 = x2/ 0.209-x

1.78 x 10-10 x 0.209 = x2

X2 = 3.7 x 10-11

therefore

x = 6.1 x 10-6M
x = [OH-]= 6.1 x 10-6M

pOH =-log [6.1 x 10-6M]= 5.21

pH = 14 – 5.21=8.79

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