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

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 :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[HCOOK]}{[HCOOH]}$

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

$pH=3.8+\log (\frac{0.3}{0.8})$

$pH=3.37$

Thus, the pH of the solution is, 3.37

Answer 2

Answer:

The pH of this aqueous solution is 3.37

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