Question

What is the pH of an aqueous solution that has 0.3 M potassium formate and 0.8 M formic acid? The pKa of formic acid is 3.8. Enter your answer to the nearest hundredth. Assume the temperature is 25 °C. Also, the value of R, the gas constant, is 1.987 cal/KLaTeX: \cdot⋅mol.

Answer 1

The pH of this aqueous solution is 3.37

Step-by-step explanation:

Step 1: Data given

Concentration of potassium formate (HCOOK) = 0.3 M

Concentration of formic acid (HCOOH) = 0.8 M

pKa of formic acid = 3.8

Step 2: Calculate pH

pH = pKa = log ([conjugate base]/[acid])

pH = 3.8 + log (0.3 M / 0.8 M)

pH = 3.8 + log (0.375)

pH = 3.37

The pH of this aqueous solution is 3.37

Answer 2

Answer : The pH of the solution is, 3.37

Explanation : Given,

`pK_a=3.8`

Concentration of
`HCOOK` = 0.3 M

Concentration of
`HCOOH` = 0.8 M

Now we have to calculate the pH of solution.

Using Henderson Hesselbach equation :

`pH=pK_a+\log ([Salt])/([Acid])`

`pH=pK_a+\log ([HCOOK])/([HCOOH])`

Now put all the given values in this expression, we get:

`pH=3.8+\log ((0.3)/(0.8))`

`pH=3.37`

Thus, the pH of the solution is, 3.37