Question

1) A 16.8 mL sample of a 0.489 M aqueous acetic acid solution is titrated with a 0.376 M aqueous sodium hydroxide solution. What is the pH at the start of the titration, before any sodium hydroxide has been added?

pH - ___

2) What is the pH at the equivalence point in the titration of of a 29.5 mL sample of a 0.460 M aqueous hydroflouric acid solution with a 0.358 M aqueous sodium hydroxide solution?

pH - ___

Answer 1

1)

Ka of CH3COOH = 1.8*10^-5

CH3COOH dissociates as:

CH3COOH -----> H+ + CH3COO-

0.489 0 0

0.489-x x x

Ka = [H+][CH3COO-]/[CH3COOH]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.8*10^-5)*0.489) = 2.967*10^-3

since c is much greater than x, our assumption is correct

so, x = 2.967*10^-3 M

So, [H+] = x = 2.967*10^-3 M

use:

pH = -log [H+]

= -log (2.967*10^-3)

= 2.5277

Answer: 2.53

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