Answer :
The [tex][OH^-][/tex] of a [tex]0. 255 M[/tex] solution of pyridine, [tex]C_5H_5N[/tex] is [tex]4.30 * 10^{-10} M[/tex].
The pH of a [tex]0. 400 M[/tex] solution of aniline, [tex]C_6H_5NH_2[/tex] is [tex]9.98.[/tex]
1] Pyridine [tex](C_5H_5N)[/tex] is a weak base, and its Kb value is given as [tex]1.69 * 10^{-9}[/tex]. To find the [tex][OH^-][/tex] of a [tex]0.255 M[/tex] solution of pyridine, we can use the following equation:
[tex]Kb = [OH^-][C_5H_5N]/[C_5H_5NH^+][/tex]
where [tex][C_5H_5NH^+][/tex]represents the concentration of the conjugate acid of pyridine, which is negligible compared to the concentration of pyridine.
Rearranging the equation and plugging in the values, we get:
[tex][OH^-] = Kb[C_5H_5N] / [C_5H_5NH^+]\\= (1.69 * 10^{-9})(0.255) / 1\\= 4.30 x 10^ {-10} M[/tex]
Therefore, the [tex][OH^-][/tex] of the solution is [tex]4.30 * 10^{-10} M.[/tex]
2] Aniline [tex](C_6H_5NH_2)[/tex] is also a weak base, and its Kb value is given as [tex]4.27 * 10^{-10}.[/tex] To find the pH of a [tex]0.400 M[/tex] solution of aniline, we can use the following equation:
[tex]Kb = [OH^-][C_6H_5NH_2]/[C_6H_5NH_3^+][/tex]
where[tex][C_6H_5NH_3^+][/tex]represents the concentration of the conjugate acid of aniline, which is formed when aniline accepts a proton from water.
We can assume that [tex]x[/tex] moles of aniline react with water to form [tex]x[/tex] moles of [tex]C_6H_5NH_3^+[/tex] and [tex]x[/tex] moles of [tex]OH^-[/tex]. Therefore, the initial concentration of aniline, [tex][C_6H_5NH_2][/tex], will decrease by [tex]x[/tex], while the concentrations of [tex][C_6H_5NH_3^+][/tex] and [tex][OH^-][/tex] will increase by [tex]x[/tex]. At equilibrium, we can express the concentrations as given follows:
[tex][C_6H_5NH_2] = 0.400 - x[/tex]
[tex][C_6H_5NH_3^+] = x[/tex]
[tex][OH^-] = x[/tex]
The value of [tex]x[/tex] can be determined using the Kb expression:
[tex]Kb = [OH^-][C_6H_5NH_2]/[C_6H_5NH_3^+]\\\\x = \sqrt{(Kb[C_6H_5NH_2]/[C_6H_5NH_3^+])} \\= \sqrt{((4.27 * 10^{-10})(0.400) / 1)}\\= 1.04 * 10^{-5} M[/tex]
Therefore, the [tex][OH^-][/tex] of the solution is[tex]1.04 * 10^{-5} M,[/tex] and the pH can be calculated using the expression:
[tex]pH = 14 - pOH\\= 14 - log([OH^-])\\= 14 - log(1.04 * 10^{-5})\\= 9.98[/tex]
Therefore, the pH of the solution is [tex]9.98[/tex].
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