Question

how many ml each are needed to make a buffer with 0.12 M HC3H5O3- (KA=1.4 *10^-4) AND 0.15 NaC3H5O3 that has a pH=3.77

Answer 1

how many ml each are needed to make a buffer with 0.12 M HC3H5O3- (KA=1.4 *10^-4) AND 0.15 NaC3H5O3 that has a pH=3.77

For given HC3H5O3/NaC3H5O3 buffer, Ka 1.4 x 10^-4, pH = 3.77,

Say M1 and V1 be the molarity and volume of salt NaC3H5O3 where M1 =0.15,

And M2, V2 be for, acid HC3H5O3 where M2 = 0.12

[NaC3H5O3] = M1V1= 0.15V1 and [HC3H5O3] = M2V2 = 0.12V2.

Henderson equation,

pH = pKa + log{[NaC3H5O3]/[ HC3H5O3]}

pH = -logKa + log{[NaC3H5O3]/[ HC3H5O3]}

3.77 = -log(1.4x10^-4) + log{[NaC3H5O3]/[ HC3H5O3]}

3.77 = 3.85 + log{[NaC3H5O3]/[ HC3H5O3]}

log{[NaC3H5O3]/[ HC3H5O3]} = 3.77 – 3.85

log{[NaC3H5O3]/[ HC3H5O3]} = 0.083

[NaC3H5O3]/[ HC3H5O3] = Antilog(0.083)

[NaC3H5O3]/[ HC3H5O3] = 1.2

0.15V1/0.12V2 =1.2

(0.15/0.12) x (V1/V2) = 1.2

V1/V2 =1.2 x (0.12 /0.15)

V1/V2 = 0.96………..(1)

i.e. V1/V2 = 96/100

i.e. 96 mL, 0.15 M NaC3H3O3 salt and 100 mL, 0.12 M HC3H5O3 acid will make 196 mL buffer with pH = 3.77.

For making 1L = 1000 mL buffer:

We have V1+V2 = 1000mL ……..(as we want to make total volume 1000 mL)

And eq.1 gives, V1 = 0.96V2. let us put it in above eqⁿ.

0.96V2 +V2 = 1000

1.96V2 =1000

V2 = 1000/1.96

V2 =510.20 mL

So, V1 = 1000-510.20 = 489.80 mL

Approximately, 510 mL salt and 490 mL acid solution of given molarity needed to be taken for making 1000 mL buffer of pH 3.77.

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