Respuesta :
Explanation:
It is known that [tex]K_{a}[/tex] of [tex]HNO_{2}[/tex] = [tex]4.5 \times 10^{-4}[/tex].
(a) Relation between [tex]K_{a}[/tex] and [tex]pK_{a}[/tex] is as follows.
[tex]pK_{a} = -log (K_{a})[/tex]
Putting the values into the above formula as follows.
[tex]pK_{a} = -log (K_{a})[/tex]
= [tex]-log(4.5 \times 10^{-4})[/tex]
= 3.347
Also, relation between pH and [tex]pK_{a}[/tex] is as follows.
pH = [tex]pK_{a} + log\frac{[conjugate base]}{[acid]}[/tex]
= [tex]3.347+ log \frac{0.15}{0.12}[/tex]
= 3.44
Therefore, pH of the buffer is 3.44.
(b) No. of moles of HCl added = [tex]Molarity \times volume[/tex]
= [tex]11.6 M \times 0.001 L[/tex]
= 0.0116 mol
In the given reaction, [tex]NO^{-}_{2}[/tex] will react with [tex]H^{+}[/tex] to form [tex]HNO_{2}[/tex]
Hence, before the reaction:
No. of moles of [tex]NO^{-}_{2}[/tex] = [tex]0.15 M \times 1.0 L[/tex]
= 0.15 mol
And, no. of moles of [tex]HNO_{2}[/tex] = [tex]0.12 M \times 1.0 L[/tex]
= 0.12 mol
On the other hand, after the reaction :
No. of moles of [tex]NO^{-}_{2}[/tex] = moles present initially - moles added
= (0.15 - 0.0116) mol
= 0.1384 mol
Moles of [tex]HNO_{2}[/tex] = moles present initially + moles added
= (0.12 + 0.0116) mol
= 0.1316 mol
As, [tex]K_{a}[/tex] = [tex]4.5 \times 10^{-4}[/tex]
[tex]pK_{a} = -log (K_{a})[/tex]
= [tex]-log(4.5 \times 10^{-4})[/tex]
= 3.347
Since, volume is both in numerator and denominator, we can use mol instead of concentration.
As, pH = [tex]pK_{a} + log \frac{[conjugate base]}{[acid]}[/tex]
= 3.347+ log {0.1384/0.1316}
= 3.369
= 3.37 (approx)
Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.
