A 1 liter solution contains 0.466 M nitrous acid and 0.350 M potassium nitrite. Addition of 0.175 moles of nitric acid will:

Question

Raise the pH slightly

Lower the pH slightly

Raise the pH by several units

Lower the pH by several units

Not change the pH

Exceed the buffer capacity

Answer 1

Answer = Lower the pH slightly. Here is the explanation:

Let us look at the composition of the original solution (before addition of the 0.175 moles of nitric acid). We have 0.466 moles nitrous acid (HNO₂) and 0.350 moles of potassium nitrite (KNO₂). HNO₂ is a weak acid, and KNO₂ is the conjugate base (salt) of that acid, so we have a BUFFER solution.

The question asks what will happen upon addition of 0.175 moles of nitric acid (HNO₃). Since this is a strong acid, it will ionize completely and so will lower the pH slightly. It will not exceed the buffer capacity because the H⁺ liberated will react with the NO₂^- present and this will form the weak acid HNO₂ and since there are 0.350 moles of NO₂^- and only 0.175 moles of HNO₃ are added, this is certainly within the buffering capacity.

We can actually calculate the original pH, and the final pH. Looking up the pKa of HNO₂, we find it to be 3.15. Using the Henderson Hasselbalch equation we can find pH of original solution:

pH = pKa + log [salt]/[acid] = 3.15 + log (0.350/0.466) = 3.15 + (-0.12) = 3.03 pH of original buffer solution.

After addition of 0.175 moles of HNO₃, concentration of conjugate base decreases by 0.175 moles and concentration of the weak acid increases by this same amount. H⁺ + NO₂^- ==> HNO₂ and assuming no change in volume upon addition of the HNO₃, we have...

Final [NO₂^-] = 0.350 - 0.175 = 0.175 M

Final [ HNO₂] = 0.466 + 0.175 = 0.641 M

pH = pKa + log [salt]/[acid] = 3.15 + log [0.175]/[0.641] = 3.15 + (-0.56) = 2.59 pH of final solution

So, the pH changed by only 0.44 pH units.