Ten Titration of Weak Acid with Strong Base A titration involves adding a reacta

Question

Ten

Titration of Weak Acid with Strong Base

A titration involves adding a reactant of known quantity to a solution of 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

HA9(aq)+OH-(aq) -----> A-(aq)+H2O(l)

A certain weak acid, HA, with a Ka value of 5.61x10-6, is titrated with NaOH

Part A

A solution is made by titrating 9.00mmol (millimole) of HA and 3.00mmol of the strong base. What is the resulting pH?

Express ther pH numerically to two decimal places.

Part B

More strong base is added until the equilance piont is reached. What is the pH of this solution at the equivalence point if the total volume is 70.0 mL?

express the pH numerically to two decimal places.

Explanation / Answer

a)

HA + OH- = A- + H2O

mmol of HA left = 9-3 = 6

mmol of A- formed = 3

pKa = -log(Ka =) -log(5.61*10^-6) = 5.25

pH = 5.25 + log(3/6) = 4.94897

b)

more bas eis added

[A-] = mmol/V = 9/70 = 0.12857

expect hydrolysis

CH2O-(aq) + H2O(l) <->CH2OH + OH-(aq)

Let HA --> CH2OH and A- = CH2O- for simplicity

since A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

The equilibrium s best described by Kb, the base constant

Kb by definition since it is an base:

Kb = [HA ][OH-]/[A-]

get ICE table:

Initially

[OH-] = 0

[HA] = 0

[A-] = M

the Change

[OH-] = + x

[HA] = + x

[A-] = - x

in Equilibrium

[OH-] = 0 + x

[HA] = 0 + x

[A-] = M - x

substitute in Kb expression

Kb = [HA ][OH-]/[A-]

(10^-14)/(5.61*10^-6) = x*x/(0.12857-x)

solve for x

x^2 + Kb*x - M*Kb = 0

solve for x with quadratic equation

x = OH- =1.5137*10^-5

[OH-]  =1.5137*10^-5

pOH = -log(OH-) = -log(1.5137*10^-5 = 4.82

pH = 14-4.82= 9.18

pH = 8.5328

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