n this question we will be calculating the points for plotting the titration cur

Question

n this question we will be calculating the points for plotting the titration curve for a weak acid using skills that we have been developing in previous homework sets (calculating pH for a weak acid solution, the common ion effect, and hydrolysis of salts).

a) You will be titrating a 1.35 M solution of C6H5COOH. What is the initial pH of this solution before the titration has begun?

b) We measure out 20.00 mL of a 1.35 M solution of C6H5COOH for our titration with sodium hydroxide. If you add 1.20 mL of a 0.90 M solution of sodium hydroxide, what is the pH of the acid solution at this stage of the titration? The Ka for C6H5COOH is 6.5 X 10-5.

((( 1.20 mL of sodium hydroxide added to the weak acid --- 0.90M sodium hydroxide in the buret as the titrant --- 20.0 mL of original 1.35 M C6H5COOH sample )))

c) If we add enough sodium hydroxide to neutralize 1/2 of our C6H5COOH, we will have converted 1/2 the C6H5COOH into C6H5COO-. At this point the [C6H5COOH] = [C6H5COO-]. What is the pH at this point that we call the half equivalence point?

d) If we add enough sodium hydroxide to neutralize all of the C6H5COOH, we will convert it all into C6H5COO- and we will have reached the equivalence point. Since the only thing that remains at the equivalence point is the salt of the conjugate base, C6H5COO-, the pH at equivalence will be determined by the hydrolysis of this salt. Since you know how many moles of C6H5COOH you started with, you know how many moles of C6H5COO- must be present at the equivalence point. To calculate the [C6H5COO-] you must know the total volume at the equivalence point. How many mL of 0.90 M sodium hydroxide are needed to reach the endpoint?

e) What is the pH at the equivalence point of this titration? Remember that since the conjugate base, C6H5COO-, is hydrolyzing in solution, you will need a Kb value for this base.

f) If we pass the equivalence point then the pH of the solution is dominated by the strong base. It is like a limiting reagent problem where now, we have run out of the weak acid and there is excess strong base. Even though there will be C6H5COO- in solution, the OH- from the strong base will control the pH. Thus, it is like a strong base pH problem. If we add 40.00 mL of sodium hydroxide we will be well past the equivalence point of the titration. What will be the pH at this point. (Hint: how much strong base will remain unreacted in solution?)

Explanation / Answer

a) C6H5COOH is a weak acid. In order to calculate pH, we first have to determine the [H+] in the solution. We can do this using the Ka equation:

Ka = [H+] [C6H5COO-] / [C6H5COOH]

Ka for C6H5COOH is 6.5 X 10-5.

Let [H+] = [C6H5COO-] = x

[C6H5COOH] = 1.35 M

Substituting the values in equation, we get

6.5 x 10-5 = x2 / 1.35

x2 = 8.78 x 10-5

x = 9.37 x 10-3

[H+] = x = 9.37 x 10-3

pH = -log [H+] = -log (9.37 x 10-3) = 2.03

b) Ka for C6H5COOH is 6.5 x 10-5.

20.00 mL of a 1.35 M solution of C6H5COOH

1.20 mL of a 0.90 M solution of NaOH

Ka for C6H5COOH is 6.5 x 10-5.

When you add NaOH to the C6H5COOH solution you form C6H5COONa. You then have a solution of a weak acid and a salt of this acid. This is a buffer solution. The pH of the buffer solution is calculated using the Henderson - Hasselbalch equation.

pH = pKa + log ([salt] /[acid])

Some preliminary calculations :

pKa = -log Ka = - log (6.5 x 10-5) = 4.19

Mol C6H5COOH in 20 mL of 1.35 M solution = (20/1000) L x 1.35 M = 0.027 mol

Mol NaOH in 1.20 mL of 0.9 M solution = (1.2/1000) L x 0.9 M = 0.00108 mol

The NaOH reacts with the C6H5COOH in 1:1 ratio, therefore you produce 0.00108 mol C6H5COONa

Remaining unreacted C6H5COOH is 0.002592 mol

In total volume 21.2 mL = 0.0212 L

Molarity of C6H5COONa = 0.00108/21.2 = 0.00005 M

Molarity of C6H5COOH = 0.002592/21.2 = 0.00012 M

Now use the H-H equation:

pH = pKa + log ([salt] /[acid])

pH = 4.19 + log (0.00005 /0.00012)

pH = 4.19 + log 0.42

pH = 4.19 + (-0.38)

pH = 3.81

(c)

Using the Henderson - Hasselbalch equation

pH = pKa + log ([salt] / [acid])

[C6H5COOH] = [C6H5COO-]

pKa = 4.19

So,

pH = pKa + log 1

pH = 4.19 + 0 = 4.19

(d)

20.00 mL of a 1.35 M solution of C6H5COOH

Mol C6H5COOH in 20 mL of 1.35 M solution = (20/1000) L x 1.35 M = 0.027 mol

0.027 mole of will neutralised by 0.027 mol of NaOH

[NaOH] = 0.9 M

Since

Molarity = Number of moles / Volume (L)

Volume (L) = Number of moles / Molarity

                    = 0.027 mol / 0.9 M

                    = 0.030 L = 30.0 mL

Hence 30 mL of 0.90 M NaOH are needed to reach the endpoint.

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