Calculate the pH of the following simple solutions: 53.1 mM HCl 0.223 M KOH 53.1 mM HCl + 0.223 M KOH Is the solution "c" a buffer? Why or why not?

Question

asked Jun 4, 2020 199k views

1 Answer

Chris Ciszak · Answer 1 · 2020-06-09T23:01:02+0000

Answer :

(1) pH = 1.27

(2) pH = 13.35

(3) The given solution is not a buffer.

Explanation :

(1) 53.1 mM HCl

Concentration of HCl =
53.1mM=53.1* 10^(-3)M

As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.

So, Concentration of hydrogen ion=
53.1* 10^(-3)M

pH : It is defined as the negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pH=-\log (53.1* 10^(-3))$

pH=1.27

(2) 0.223 M KOH

Concentration of KOH = 0.223 M

As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.

So, Concentration of hydroxide ion= 0.223 M

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (0.223)$

pOH=0.65

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.65=13.35$

(3) 53.1 mM HCl + 0.223 M KOH

Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.

It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.

As we know that the pH of strong acid and strong base solution is always 7.

So, the given solution is not a buffer.