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

Answer:

The correct option is C

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

       $K_c = \frac{[H^+] [A^-]}{[HA]}$

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  

        $K_c = \frac{0.0031623 * 0.0031623 }{(2.5 - 0.0031623 }$

         $K_c = 4* 10^{-6}$