Answer :
Answer: The pH at equivalence for the given solution is 8.37.
Explanation:
We know that,
[tex]pK_{a} = -log K_{a}[/tex]
3.46 = [tex]-log K_{a}[/tex]
[tex]K_{a} = 3.467 \times 10^{-4}[/tex]
Now, we will calculate the volume of NaOH required to reach the equivalence point as follows.
[tex]M_{(HCNO)} \times V_{(HCNO)} = M_{(NaOH)} \times V_{(NaOH)}[/tex]
[tex]0.3065 M \times 160.0 mL = 0.4994M \times V_{(NaOH)}[/tex]
[tex]V_{(NaOH)}[/tex] = 98.1978 mL
The given data is as follows.
M(HCNO) = 0.3065 M
V(HCNO) = 160 mL
M(NaOH) = 0.4994 M
V(NaOH) = 98.1978 mL
So, moles of HCNO will be calculated as follows.
mol(HCNO) = [tex]M(HCNO) \times V(HCNO)[/tex]
mol(HCNO) = [tex]0.3065 M \times 160 mL[/tex]
= 49.04 mmol
Now, the moles of NAOh will be calculated as follows.
mol(NaOH) = [tex]M(NaOH) \times V(NaOH)[/tex]
mol(NaOH) = [tex]0.4994 M \times 98.1978 mL[/tex]
= 49.04 mmol
This means that, 49.04 mmol of both HCNO and NaOH will react to form [tex]CNO^{-}[/tex] and [tex]H_{2}O[/tex].
Here, [tex]CNO^{-}[/tex] is strong base . So,
[tex]CNO^{-}[/tex] formed = 49.04 mmol
Total volume of the solution is as follows.
Volume of Solution = 160 + 98.1978 = 258.1978 mL
And,
[tex]K_{b}[/tex] of [tex]CNO^{-}[/tex] = [tex]\frac{K_{w}}{K_{a}}[/tex]
= [tex]\frac{1 \times 10^{-14}}{3.467 \times 10^{-4}}[/tex]
= [tex]2.884 \times 10^{-11}[/tex]
The concentration of [tex]CNO^{-}[/tex] is as follows.
c = [tex]\frac{49.04 mmol}{258.1978 mL}[/tex]
= 0.1899M
Also,
[tex]CNO^{-} + H_{2}O \rightarrow HCNO + OH^{-}[/tex]
Initial: 0.1899 0 0
Equilibm:0.1899 - x x x
[tex]K_{b} = \frac{[HCNO][OH^{-}]}{[CNO^{-}]}[/tex]
[tex]K_{b} = \frac{x \times x}{(c - x)}[/tex]
We assume that x can be ignored as compared to c . Hence, above formula can be rewritten as follows.
[tex]K_{b} = \frac{x \times x}{c}[/tex]
so, x = [tex]\sqrt (K_{b} \times c)[/tex]
x = [tex]\sqrt ((2.884 \times 10^{-11}) \times 0.1899)[/tex]
= [tex]2.34 \times 10^{-6}[/tex]
Here, c is much greater than x, this means that our assumption is correct .
so, x = [tex]2.34 \times 10^{-6}[/tex] M
[tex][OH^{-}] = x = 2.34 \times 10^{-6}[/tex] M
As,
pOH = [tex]-log [OH^{-}][/tex]
= [tex]-log (2.34 \times 10^{-6})[/tex]
= 5.6307
Also, pH = 14 - pOH
= 14 - 5.6307
= 8.3693
or, = 8.37 (approx)
Thus, we can conclude that the pH at equivalence for the given solution is 8.37.