A volume of 500.0 mL of 0.160 M NaOH is added to 585 mL of 0.200 M weak acid ( K a = 1.28 × 10 − 5 ) . What is the pH of the resulting buffer? HA ( aq ) + OH − ( aq ) ⟶ H 2 O ( l ) + A − ( aq

Answer : The pH of the resulting buffer is, 5.22

Explanation : Given,

First we have to calculate the moles of

and,

The balanced chemical reaction is:

Moles of HA left = 0.117 mol - 0.08 mol = 0.037 mol

Moles of = 0.08 mol

The expression used for the calculation of is,

Now put the value of in this expression, we get:

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

Now put all the given values in this expression, we get:

Thus, the pH of the resulting buffer is, 5.22

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