When a strong base is gradually added drop wise to a weak acid, the pH changes a

Question

When a strong base is gradually added drop wise to a weak acid, the pH changes at each addition. When the appropriate quantity of base has been added to react with all of the acid, the pH changes sharply, indicating the endpoint of the titration. A plot of pH versus volume of base added gives what is known as a titration curve. Consider the titration of 25.00 mL of 0.1000 M acetic acid with 0.1000 M NaOH. a. Write the equation for the titration reaction. b. Determine the volume of NaOH solution required to reach the endpoint. c. The chart below has entries for several steps along the titration curve. To calculate the pH at each step, you first must understand the chemistry at that step, and then you can decide the appropriate method to calculate the pH. For each volume listed, (i) list the major species in solution, (ii) determine whether the K_a equation, K_b equation, buffer equation, or solution equilibrium equation is appropriate for the calculation of the solution [H^+] and pH, and (iii) complete the calculations.

Explanation / Answer

a) Equation for titration reaction

CH3COOH (aq) + NaOH (aq) ---------à CH3COONa (aq) + H2O(l)

b) Calculate the volume of NaOH needed to reach the equivalence point

Given

0.1000M NaOH

Volume of CH3COOH = 50 ml,0.1000 M acetic acid

Moles CH3COOH = moles NaOH

Ma × Va = Mb × Vb

Vb = Ma Va /Mb = (0.1000M)(50.0mL)/0.1000M = 50.0 mL

c)

Volume of NaOH is 0

Before adding NaOH the pH is that for a solution of 0.1000 M acetic acid. Because acetic acid is a weak acid,

CH3COOH (aq)+H2O(l) H3O+(aq)+CH3COO(aq)

Ka = [H3O+][CH3COO]/[CH3COOH]= (x) (x) /0.100x = 1.75×105

X = [H3O+] = 1.32×103M

PH = - log [H3O+] = - log 1.32×103 = 2.88

Volume of NaOH is 5ml

Adding 5 ml NaOH converts a portion of the acetic acid to its conjugate base, CH3COO–.

CH3COOH (aq) +OH(aq)H2O(l)+CH3COO(aq)

[CH3COOH] = (50ml) (0.1000M) – (5ml) (0.1000M) / (5ml) + 50ml) = 0.0818M

[CH3COO-] = (5ml) (0.1000M) / (5ml) + (50ml) = 0.0090M

PKa = 4.76

PH = PKa + log 0.0090M / 0.0818M

      = 4.76 - 0.96 = 3.80

Volume of NaOH is 12.5ml

[CH3COOH] = (50ml) (0.1000M) – (12.5ml) (0.1000M) / (12.5ml) + 50ml) = 0.06M

[CH3COO-] = (12.5ml) (0.1000M) / (12.5ml) + (50ml) = 0.02M

PH = PKa + log 0.02M / 0.06M

      = 4.76 - 0.477 = 4.283

Volume of NaOH is 20ml

[CH3COOH] = (50ml) (0.1000M) – (20ml) (0.1000M) / (20ml) + 50ml) = 0.0428M

[CH3COO-] = (20ml) (0.1000M) / (20ml) + (50ml) = 0.0285M

PH = PKa + log 0.0285M / 0.0428M

      = 4.76 - 0.168 = 4.592

Volume of NaOH is 25ml

[CH3COOH] = (50ml) (0.1000M) – (25ml) (0.1000M) / (25ml) + 50ml) = 0.0333M

[CH3COO-] = (25ml) (0.1000M) / (25ml) + (50ml) = 0.0333M

PH = PKa + log 0.0333M / 0.0333M

    = 4.76 - 0.0 = 4.76

Volume of NaOH is 30ml

[CH3COOH] = (50ml) (0.1000M) – (30ml) (0.1000M) / (30ml) + 50ml) = 0.025M

[CH3COO-] = (30ml) (0.1000M) / (30ml) + (50ml) = 0.0375M

PH = PKa + log 0.0375M / 0.025M

    = 4.76 + 0.176 = 4.936

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