Respuesta :
Answer: The pH after 0.02 mol of [tex]Ba(OH)_{2}[/tex] are added to 0.72 L of the solution is 9.5.
Explanation:
The chemical equation for the dissociation of base represented by B in water is depicted as follows.
[tex]B(aq) + H_{2}O(l) \rightleftharpoons BH^{+}(aq) + OH^{-}(aq)[/tex]
According to Henderson-Hasselbach equation,
pH = [tex]pK_{a} + log \frac{[B]}{[BH^{+}]}[/tex]
9.31 = [tex]pK_{a} + log \frac{0.17}{0.39}[/tex]
[tex]pK_{a}[/tex] = 9.31 + 0.361
= 9.671
Initial moles of B is as follows.
[tex]0.72 L \times \frac{0.17 M}{L}[/tex]
= 0.1224 mol B
Now, the initial concentration of [tex]BH^{+}[/tex] is as follows.
[tex]0.72 L \times \frac{0.39 mol}{L}[/tex]
= 0.281 mol [tex]BH^{+}[/tex]
We will calculate the equilibrium moles after the addition of 0.02 moles [tex]Ba(OH)_{2}[/tex] which is 0.04 moles [tex]OH^{-}[/tex] as follows.
[tex]BH^{+}(aq) + OH^{-}(aq) \rightleftharpoons B(aq) + H_{2}O(aq)[/tex]
Initial: 0.281 0.04 0.1224
Change: -0.04 -0.04 +0.04
Equilbm: 0.241 0 0.1624
Hence,
pH = [tex]pK_{a} + log \frac{[B]}{[BH^{+}]}[/tex]
= [tex]9.671 + log \frac{0.1624/0.72}{0.241/0.72}[/tex]
= [tex]9.671 + log (0.225/0.334)[/tex]
= 9.671 - 0.171
= 9.5
Thus, we can conclude that the pH after 0.02 mol of [tex]Ba(OH)_{2}[/tex] are added to 0.72 L of the solution is 9.5.
