20 mL of 0.1 M acetic acid is mixed with 50 mL of potassium acetate. Ka of acetic acid =1.8 × 10-5 at 27° C. Calculate concentration of potassium acetate if pH of the mixture is 4.8.

Question

20 mL of 0.1 M acetic acid is mixed with 50 mL of potassium acetate. Ka of acetic acid =1.8 × 10-5 at 27° C. Calculate concentration of potassium acetate if pH of the mixture is 4.8. (A) 0.1 M (B) 0.04 M (C) 0.4 M (D) 0.02 M

Tardigrade Admin · Accepted Answer

Let the concentration of potassium acetate is

x

.
According to Henderson's equation,

p H = p K_{a} + lo g \frac{[ salk ]}{[ acid ]}

4.8 = - lo g (1.8 \times 1 0^{- 5}) + lo g \frac{x \times 50}{20 \times 0.1 M}

4.8 = 4.74 + lo g 25 x

or

lo g 25 x = 0.06

25 x = 1.148

∴ x = 0.045 M

Q. 20mL of 0.1M acetic acid is mixed with 50mL of potassium acetate. Ka​ of acetic acid =1.8×10−5 at 27∘C. Calculate concentration of potassium acetate if pH of the mixture is 4.8.

0.1 M

0.04 M

0.4 M

0.02 M

Let the concentration of potassium acetate is x. According to Henderson's equation, pH=pKa​+log[acid][ salk]​ 4.8=−log(1.8×10−5)+log20×0.1Mx×50​ 4.8=4.74+log25x or log25x=0.06 25x=1.148 ∴x=0.045M

Q. $20 m L$ of $0.1 M$ acetic acid is mixed with $50 m L$ of potassium acetate. $K_{a}$ of acetic acid $= 1.8 \times 1 0^{- 5}$ at $2 7^{\circ} C$ . Calculate concentration of potassium acetate if $p H$ of the mixture is $4.8$ .

Let the concentration of potassium acetate is $x$ .
According to Henderson's equation,
$p H = p K_{a} + lo g \frac{[ salk ]}{[ acid ]}$
$4.8 = - lo g (1.8 \times 1 0^{- 5}) + lo g \frac{x \times 50}{20 \times 0.1 M}$
$4.8 = 4.74 + lo g 25 x$
or $lo g 25 x = 0.06$
$25 x = 1.148$
$∴ x = 0.045 M$