Fluoroacetic acid occurs in gifblaar, one of the most poisonous of all plants. A

Question

Fluoroacetic acid occurs in gifblaar, one of the most poisonous of all plants. A 0.318 M solution of the acid is found to have a pH =1.56. CH_2FCOOH (aq) + H_20 (1) doubleheadarrow H_3o^+(aq) + CH_2FCOO i) Calculate the Ka of fluoroacetic acid. ii) Calculate the new pH if 2.1 g of sodium fluoroacetate (CH_2FCOONa) is added to 100 ml of the 0.318 M fluoroacetic acid solution from part i). iii) Calculate the new pH if 0.05 g of NaQH (s) is added to the fluoroacetic acid /fluoracetate solution from part ii).

Explanation / Answer

Assume that; fluoroacetic acid be denoted as HFA.

Then HFA===> H+ + FA-

Ka = [H+][FA-]/[HFA]

Given that; pH = 1.56

pH = -Log[H+], and Log[H+] =10^ -1.56.
= 2.75 x 10^-2

[H+] = [FA-] = 2.75 x 10^-2

And HF A= 0.318 now calculate the Ka as follows:

Ka = (2.75x10^-2)(2.75x10^-2)/(0.318)

= 2.38 x 10^-3

To calculate the pH of buffer solution use Use Henderson - Hasselbalch equation

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

sodium flouro acetate = 100 g/mol

2.1 g / 100 g/mol = 0.021 moles sodium flouro acetate

[salt]= 0.021/0.100= 0.21 M

Moles of flouroacetic acid = 0.318 *0.100= 0.0318 moles

[acid] = 0.0318/0.100= 0.318 M

pKa= -log Ka= -log 2.38 x 10^-3

= 2.62

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

pH = 2.62 + log 0.21/0.318

pH = 2.62-0.180

=2.44

Calculate the moles of NaOH in 0.05 g as follows:

Molar mass of NaOH = 39.997 g/mol

0.05/ 39.997 g/mol= 1.25*10^-3 moles

Moles of flouroacetic acid ; HFA= 0.318 *0.100= 0.0318 moles

0.021= A-; moles sodium flouro acetate

moles NaOH added = 1.25*10^-3 moles

HFA + OH- => FA- + H2O
Now new moles of acid and salt :

moles HFA = 0.0318 - 1.25*10^-3 =0.03055
moles FA- = 0.021 + 1.25*10^-3 = 0.02225

total volume = 0.100 L

[HFA]= 0.03055/ 0.100=0.3055 M
[FA-]= 0.02225 /0.100 =0.2225 M

pH = 2.62 + log 0.2225/ 0.3055

pH = 2.62 -0.138

=2.482

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