Respuesta :
Explanation:
The given reaction is as follows.
[tex]HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)[/tex]
Hence, number of moles of NaOH are as follows.
n = [tex]0.05 L \times 0.1 M[/tex]
= 0.005 mol
After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.
n = [tex]0.025 L \times 0.1 M[/tex]
= 0.0025 mol
According to ICE table,
[tex]HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)[/tex]
Initial: 0.005 mol 0.0025 mol 0 0
Change: -0.0025 mol -0.0025 mol +0.0025 mol
Equibm: 0.0025 mol 0 0.0025 mol
Hence, concentrations of HA and NaA are calculated as follows.
[HA] = [tex]\frac{0.0025 mol}{V}[/tex]
[NaA] = [tex]\frac{0.0025 mol}{V}[/tex]
[tex][A^{-}] = [NaA] = \frac{0.0025 mol}{V}[/tex]
Now, we will calculate the [tex]pK_{a}[/tex] value as follows.
pH = [tex]pK_{a} + log \frac{A^{-}}{HA}[/tex]
[tex]pK_{a} = pH - log \frac{[A^{-}]}{[HA]}[/tex]
= [tex]3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}[/tex]
= 3.42
Thus, we can conclude that [tex]pK_{a}[/tex] of the weak acid is 3.42.
The pKa of the weak acid is 3.42.
Given reaction:
HA +NaOH → NaA+ H₂O
Hence, number of moles of NaOH are as follows.
n =0.05L * 0.1 M
n= 0.005 mol
After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.
n = 0.025L * 0.1 M
n = 0.0025 mol
According to ICE table,
Initial: 0.005 mol 0.0025 mol 0
Change: -0.0025 mol -0.0025 mol +0.0025 mol
Equibm: 0.0025 mol 0 0.0025 mol
Hence, concentrations of HA and NaA are calculated as follows.
[HA] = 0.0025 mol / V
[tex][A^-][/tex] = [NaA] = 0.0025 mol / V
Calculation for pKa:
[tex]pH = pKa + log \frac{A^-}{HA} \\\\pKa = 3.42 + log 1\\\\pKa =3.42[/tex]
Thus, we can conclude that pKa of the weak acid is 3.42.
Find more information about pKa here:
brainly.com/question/22390063
