Question

what is the ph of a solution of 0.400 m ch₃nh₂ containing 0.250 m ch₃nh₃i? (k_b of ch₃nh₂ is 4.4 × 10⁻⁴)

Explanation:

Step1: Write the reaction equation

$$\ce{CH3NH2 + H2O <=> CH3NH3+ + OH-}$$
Let \(x\) be the concentration of \(\ce{OH-}\) and \(\ce{CH3NH3+}\) at equilibrium. The concentration of \(\ce{CH3NH2}\) at equilibrium is \(0.400 - x\). Since \(K_b\) is small, \(x\) is negligible compared to \(0.400\), so \(0.400 - x\approx0.400\).

Step2: Use the \(K_b\) formula

\(K_b=\frac{[\ce{CH3NH3+}][\ce{OH-}]}{[\ce{CH3NH2}]}\)
Given \(K_b = 4.4\times10^{-4}\), substituting the values:
\(4.4\times10^{-4}=\frac{x\cdot x}{0.400}\)
\(x^{2}=4.4\times10^{-4}\times0.400\)
\(x^{2}=1.76\times 10^{-4}\)
\(x = [\ce{OH-}]=\sqrt{1.76\times 10^{-4}}\approx1.33\times10^{-2}\space M\)

Step3: Calculate \(pOH\)

\(pOH=-\log([\ce{OH-}])=-\log(1.33\times 10^{-2})\)
\(pOH\approx1.88\)

Step4: Calculate \(pH\)

Using the formula \(pH + pOH=14\)
\(pH=14 - pOH\)
\(pH=14 - 1.88 = 12.12\)

Answer:

\(12.12\)