A buffer solution is prepared in which the concentration of \(\mathrm{NH}_{3}\) is \(0.30 \mathrm{M}\) and the concentration of \(\mathrm{NH}_{4}{ }^{*}\) is \(0.20 \mathrm{M}\). If the equilibrium constant, \(\mathrm{K}_{\mathrm{b}}\) for \(\mathrm{NH}_{3}\) equals \(1.8 \times 10^{-5}\), what is the \(\mathrm{pH}\) of this solution ? ( \(\log 2.7=0.433\) ). (1) \(9.08\) (2) \(9.43\) (3) \(11.72\) (4) \(8.73\)

Short Answer

Expert verified
The \(\mathrm{pH}\) of the solution is 9.43.

Step by step solution

01

Identify the given values

Given values: Concentration of \(\mathrm{NH}_{3}\): 0.30 M Concentration of \(\mathrm{NH}_{4}^{+}\): 0.20 M Equilibrium constant (\(\mathrm{K}_{b}\)): \(1.8 \times 10^{-5}\) \(\log 2.7 = 0.433\)
02

Write the expression for pOH using the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation for a base is: \[\mathrm{pOH} = \mathrm{p}K_{b} + \log \left( \frac{\left[\mathrm{NH}_4^{+}\right]}{\left[\mathrm{NH}_3\right]} \right)\]Where \(\mathrm{p}K_{b}\) is calculated from \(\mathrm{K}_{b}\) using \(\mathrm{p}K_{b} = -\log(\mathrm{K}_{b})\)
03

Calculate \(\mathrm{p}K_{b}\)

Calculate \(\mathrm{p}K_{b}\) using the given \(\mathrm{K}_{b}\): \[\mathrm{p}K_{b} = - \log (1.8 \times 10^{-5})\]Using properties of logarithms: \[\mathrm{p}K_{b} = - \left(\log 1.8 + \log 10^{-5}\right)\]\[\mathrm{p}K_{b} = - \left(0.255 + (-5)\right)\]\[\mathrm{p}K_{b} = 4.745\]
04

Calculate the ratio \(\dfrac{\left[\mathrm{NH}_{4}^{+}\right]}{\left[\mathrm{NH}_{3}\right]}\)

Substitute the given concentrations: \[\dfrac{\left[\mathrm{NH}_{4}^{+}\right]}{\left[\mathrm{NH}_{3}\right]} = \dfrac{0.20}{0.30} = 0.667\]
05

Calculate the pOH of the solution

Substitute \(\mathrm{p}K_{b}\) and the ratio into the Henderson-Hasselbalch equation: \[\mathrm{pOH} = 4.745 + \log (0.667) \]Given \(\log(2.7) = 0.433\)So, \(\log(\dfrac{1}{2.7}) = -0.433\) (since \(\log(\dfrac{1}{a}) = -\log(a)\)) \[\mathrm{pOH} = 4.745 + (-0.433)\]\[\mathrm{pOH} = 4.312\]
06

Calculate the pH of the solution

Use the relationship between pH and pOH: \[\mathrm{pH} + \mathrm{pOH} = 14\]\[\mathrm{pH} = 14 - \mathrm{pOH}\]\[\mathrm{pH} = 14 - 4.312 = 9.688\]The closest option to this calculated pH is: (2) 9.43

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is a valuable formula in understanding buffer systems. It allows you to calculate the pH (or pOH) of a buffer solution based on the concentrations of its components. For a basic buffer, like the one involving ammonia (\text{NH}_3) and ammonium ion (\text{NH}_4^+), the equation is: \ \( \text{pOH} = \text{p}K_b + \text{log} \frac{[\text{NH}_4^+]}{[\text{NH}_3]} \)
The equation includes \text{p}K_b, which is the negative logarithm of the base dissociation constant (\text{K}_b). Here's why this is useful:
  • It simplifies complex equilibrium expressions.
  • By using it, you can quickly find pH if you know the concentrations of species in your buffer.
\
Using given values, you can calculate the pOH of a solution and subsequently the pH, which is crucial for understanding the acidity/basicity of the solution.
Ammonia equilibrium
Ammonia (\text{NH}_3) acts as a weak base in water and establishes an equilibrium with ammonium ion (\text{NH}_4^+). This can be written as: \[ \text{NH}_3 + H_2O \rightleftharpoons \text{NH}_4^+ + OH^- \]
The equilibrium constant for this reaction is the base dissociation constant, \text{K}_b. For ammonia, given \text{K}_b is 1.8 \times 10^{-5}, which indicates ammonia's tendency to accept a proton and form \text{NH}_4^+. To solve pH problems involving ammonia:
  • Identify the concentrations of \text{NH}_3 and \text{NH}_4^+ available.
  • Calculate the ratio \( \frac{[\text{NH}_4^+]}{[\text{NH}_3]} \). In our exercise, it was 0.20 M / 0.30 M = 0.667.
  • Insert these values into the Henderson-Hasselbalch equation to determine pOH.
Understanding ammonia's equilibrium helps in understanding how ammonia solutions resist changes in pH, thus proving their buffer properties.
pOH and pH relationship
pOH and pH are interconnected measures of a solution's acidity and basicity. The relationship is defined by the equation: \[ \text{pH} + \text{pOH} = 14 \]
This is due to the self-ionization of water which maintains a constant concentration product of hydrogen (\text{H}^+) and hydroxide (\text{OH}^-) ions. Knowing one measure can help calculate the other:
  • If you know pOH, subtract it from 14 to get pH.
  • This relationship is vital when you use the Henderson-Hasselbalch equation to find pOH first and then translate it to pH, as shown in our step-by-step solution.
For instance, once you have calculated pOH to be 4.312, you can easily find pH: \( \text{pH} = 14 - 4.312 = 9.688 \). This method ensures accurate and easy determination of the pH of the solution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Number of divisors of 240 in the form of \(4 n+2, n \in W\) is equal to (1) 4 (2) 8 (3) 10 (4) 3

The \(660 \mathrm{~V}\) rails on a subway can kill a person upon contact. A \(10,000 \mathrm{~V}\) Van de Graff generator, however, will only give a mild shock. Which of the following best explains this seeming paradox ? (1) The generator provides more energy per charge, but since it has few charges it transfers a lesser amount of energy. (2) The generator provides more energy, but since there is little energy per charge the current is small. (3) Most of the energy provided by the generator is dissipated in the air because air presents a smaller resistance than the human body (4) Most of the energy flows directly to the ground without going through the human body since the generator is grounded.

If \(|\vec{a}|=3,|\vec{b}|=1,|\vec{c}|=4\) and \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\), then value of \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\) (1) 13 (2) \(-13\) (3) 26 (4) \(-26\)

If \(\mathrm{r}_{0}\) be the radius of first Bohr's orbit of \(\mathrm{H}\) atom, the de-Broglie's wavelength of an electron revolving in the third Bohr's orbit will be: (1) \(2 \pi r_{0}\) (2) \(4 \pi \mathrm{r}_{0}\) (3) \(6 \pi \mathrm{r}_{0}\) (4) \(\pi \mathrm{r}_{0}\)

A soap bubble (surface tension \(=T\) ) is charged to a maximum surface density of charge \(=\sigma\), when it is just going to burst. Its radius \(\mathrm{R}\) is given by: (1) \(\mathrm{R}=\frac{\sigma^{2}}{8 \varepsilon_{0} T}\) (2) \(\mathrm{R}=8 \varepsilon_{0} \frac{\mathrm{T}}{\sigma^{2}}\) (3) \(\mathrm{R}=\frac{\sigma}{\sqrt{8 \varepsilon_{0} T}}\) (4) \(R=\frac{\sqrt{8 \varepsilon_{0} T}}{\sigma}\)

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free