A buffer solution contains 0.475 M nitrous acid and 0.302 M sodium nitrite . If 0.0224 moles of potassium hydroxide are added to 150 mL of this buffer, what is the pH of the resulting solution

The pH of the resulting solution will be 3.001

Explanation:

Molarity is calculated by using the equation:

         ......(1)

We are given:

Moles of NaOH = 0.0224 moles

Molarity of nitrous acid = 0.475 M

Molarity of sodium nitrite = 0.302 M

Volume of solution = 150 mL = 0.150 L          (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

The chemical equation for the reaction of nitrous acid and NaOH follows:

I:                0.07125     0.0224     0.0453

C:             -0.0224     -0.0224   +0.0224

E:              0.04885         -          0.0677

The power of the acid dissociation constant is the negative logarithm of the acid dissociation constant. The equation used is:

       ......(2)

We know:

for nitrous acid =

Using equation 2:

To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:

        .......(3)

Given values:

Putting values in equation 3. we get:

Hence, the pH of the resulting solution will be 3.001

i belive it is water