Chapter 19: Q19.20P (page 873)

Find the pH of a buffer that consists of 0.50M methylamine ( ${CH}_{3} {NH}_{2}$ ) and 0.60M ${CH}_{3} {NH}_{3} Cl$ ( ${pK}_{b}$ of ${CH}_{3} {NH}_{2} = 3.35$ )?

Short Answer

Expert verified

The pH value is 10.57.

Step by step solution

Basic buffer solution

The combination of a weak base with a salt of its conjugate acid forms a basic buffer solution. The mixture of methylamine and methyl ammonium chloride forms a basic buffer solution.

The pH of the solution

Write the 0.50 M ${CH}_{3} {NH}_{2}$ dissociation equation first.

${CH}_{3} {NH}_{2} {+H}_{2} O ⇌ {CH}_{3} N {H_{3}}^{+} {+ OH}^{-} .$

Then write the 0.60 M ${CH}_{3} {NH}_{3} Cl$ dissociation equation.

${CH}_{3} {NH}_{3} Cl \to {CH}_{3} N {H_{3}}^{+} {+ Cl}^{-} .$

We now have the initial methylamine and methyl ammonium concentrations.

${CH}_{3} {NH}_{2} = 0.50 M. {CH}_{3} N {H_{3}}^{+} = 0.60 M.$

To find the pH, we may utilize the Henderson-Hasselbalch equation. First, solve for the value of ${pK}_{a}$ as:

${pK}_{w} {= pK}_{a} {+ pK}_{b} {pK}_{a} {= pK}_{w} {+ pK}_{b} = 14 - 3.35 = 10.65. pH = pKa + log \frac{[base]}{[acid]} = pKa + log \frac{[{CH}_{3} {NH}_{2}]}{[{CH}_{3} N {H_{3}}^{+}]} = 10.65 + log \frac{0.50 M}{0.60 M} = 10.57.$

Therefore, the pH value is 10.57.

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Find the pH of a buffer that consists of 0.50M methylamine ( ${CH}_{3} {NH}_{2}$ ) and 0.60M ${CH}_{3} {NH}_{3} Cl$ ( ${pK}_{b}$ of ${CH}_{3} {NH}_{2} = 3.35$ )?

Short Answer

Step by step solution

Basic buffer solution

The pH of the solution

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Find the pH of a buffer that consists of 0.50M methylamine (CH3NH2) and 0.60M CH3NH3Cl(pKbofCH3NH2= 3.35)?

Short Answer

Step by step solution

Basic buffer solution

The pH of the solution

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Making Measurements

The Earths Atmosphere

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Study anywhere. Anytime. Across all devices.

Find the pH of a buffer that consists of 0.50M methylamine ( ${CH}_{3} {NH}_{2}$ ) and 0.60M ${CH}_{3} {NH}_{3} Cl$ ( ${pK}_{b}$ of ${CH}_{3} {NH}_{2} = 3.35$ )?