Question

A 0.270 M monoprotic weak acid solution has a pH of 2.50. What is the pKa of this acid? Select one: a. 4.43 b. 1.93 C. 5.57 d. 9.57 e. 8.07

Answer 1

Given,

Molarity of weak acid (HA) = 0.270M

pH = 2.50

pKa = ?

We know that, dissociation equation for acid is

HA  $\rightleftharpoons$ H⁺ + A^-

Also,

pH = -log[H⁺]

So, first we will calculate the concentration of H⁺ using pH

Put the pH in above formula,

pH = -log[H⁺]

2.50 = -log[H⁺]

[H⁺] = 10^-2.50

[H⁺] = 0.00316 M

Now, From the reaction above we can conclude that there is a 1:1 molar ratio between [H⁺] and [A¯].

So, Concentration of [A¯] is same as [H⁺]

[A¯] = 0.00316 M

Now, we know also know that,

pK_a = pH + log₁₀([conjugate acid]/[conjugate base])

pK_a = pH + log₁₀([HA]/[A^-])

put all the known values,

pK_a = 2.50 + log₁₀(0.270/0.00316)

pK_a = 2.50 + log₁₀(85.44)

pK_a = 2.50 + 1.932

pK_a = 4.432

So, OPTION A (4.43) is Correct.