Question

Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (NaCNO) in enough water to make 1.0 liter of solution. [K a(HCNO) = 2.0 × 10 −4] 0.97 3.10 4.40 3.70 4.30

Answer 1

Answer: Option (e) is the correct answer.

Explanation:

Equilibrium reaction equation for the given reaction is as follows.

    $HCNO(aq) + H_{2}O(l) \rightarrow CNO^{-}(aq) + H_{3}O^{+}(aq)$

It is given that initial moles of HCNO is 0.20 mol and for NaCNO is 0.80 mol. $K_{a}$ of HCNO is $2 \times 10^{-4}$ mol.

Now, we will assume that at equilibrium there are x moles.

            $HCNO(aq) + H_{2}O(l) \rightarrow CNO^{-}(aq) + H_{3}O^{+}(aq)$

Initial:        0.20                           0.80              0

Change:      -x                              +x                +x

Equilibrium: 0.20 - x                 0.80 + x           x

As the volume of the given solution is 1 liter, equilibrium concentration and moles are same.

           $K_{a} = \frac{[CNO^{-}][H_{3}O^{+}]}{[HCNO]}$

       $2.0 \times 10^{-4} = \frac{x(0.80 + x)}{(0.20 - x)}$

              x = $5.0 \times 10^{-5}$ M

Then, pH = $-log[H_{3}O^{+}]$

               = $-log (5.0 \times 10^{-5})$

               = 4.30

Thus, we can conclude that pH of given buffer solution is 4.30.