Answer:
[tex]\boxed{\text{10.84}}[/tex]
Explanation:
A solution of a weak base and its conjugate acid is a buffer.
The equation for the equilibrium is
[tex]\rm CH$_3$NH$_2$ + H$_2$O $\, \rightleftharpoons \,$ CH$_3$NH$_2$+ H$_{3}$O$^{+}$\\\text{For ease of typing, let's rewrite this equation as}\\\rm B + H$_2$O $\longrightarrow \,$ BH$^{+}$ + OH$^{-}$; $K_{\text{b}}$ = 4.4 \times 10^{-4}$[/tex]
The Henderson-Hasselbalch equation for a basic buffer is
[tex]\text{pOH} = \text{p}K_{\text{b}} + \log\dfrac{[\text{BH}^{+}]}{\text{[B]}}[/tex]
Data:
[B] = 0.400 mol·L⁻¹
[BH⁺] = 0.250 mol·L⁻¹
Kb = 4.4 × 10⁻⁴
Calculations:
(a) Calculate pKb
pKb = -log(4.4× 10⁻⁴) = 3.36
(b) Calculate the pH
[tex]\text{pOH} = 3.36 + \log \dfrac{0.250}{0.400} = 3.36 + \log 0.625 = 3.36 - 0.204 = 3.16\\\\\text{pH} =14.00 -3.16 = \mathbf{10.84}\\\\\text{The pH of the solution is }\boxed{\textbf{10.84}}[/tex]
The pH of a solution of 0.400 M CH₃NH₂ containing 0.250 M CH₃NH₃I = 10.84
This given solution contains a weak base (CH₃NH₂) and the conjugate acid of that weak base (CH₃NH₃I) which makes a Buffer.
We know that
[tex]pOH=pK_b+log\frac{[salt]}{[base]}[/tex]
or, [tex]pH=pK_a+log\frac{[base]}{[salt]}[/tex]
CH₃NH₂ is the base and CH₃NH₃I is the corresponding conjugate acid (salt).
Given:
[base] = 0.400 M
[salt] or [acid] = 0.250 M
= 4.4 * 10-4
So = 3.36
solution:
Putting in the equation,
[tex]pOH= 3.36 +log\frac{0.250\ M}{0.400\ M}[/tex]
pOH= 3.16
So pH = 14 - pOH
= 10.84
Thus, the pH of a solution of 0.400 M CH₃NH₂ containing 0.250 M CH₃NH₃I = 10.84
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