Question

A 2.5 M solution of a monoprotic acid has a pH of 2.50 at 25°C. What is the equilibrium constant for this acid? Is it a strong or weak acid?

A) 128, strong
B) 2.44, strong
C) 4.1×10−6, strong
D) 4.1×10−6, weak

Answer 1

The correct option is C

Step-by-step explanation:

From the question we are told that

The concentration of the acid solution is
`C = 2.5M`

The pH of the acid is
`pH_a = 2.5`

Looking at the pH of the acid we can say that it is a strong acid

This because its pH fall between 0-7

The generally dissociation chemical equation for this acid is represented as

`HA` ⇄
`H^(+) + A^(-)`

So the equilibrium constant for this reaction is mathematically represented as

`K_c = ([H^+] [A^-])/([HA])`

and the pH of can be mathematically represented as

`pH = -log [H^+]`

=>
`H^+ = 10^(-pH)`

substituting value

`H^+ = 10^(-(2.50))`

`H^+ = 0.0031623M`

Looking at the chemical equation we see that the ratio of the moles for the product is one is to one

Secondly the new concentration of the reactant will be

`HA = 2.5 - 00031623`

`[A^-] = 0.003163 M`

So the equilibrium constant is

`K_c = (0.0031623 * 0.0031623 )/((2.5 - 0.0031623 )`

`K_c = 4* 10^(-6)`