Question

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?

Answer 1

(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.