The \(\mathrm{p} K_{\mathrm{b}}\) of water is ______ . (a) 1 (b) 7 (c) 14 (d) not defined (e) none of the above

Short Answer

Expert verified
The $\mathrm{p}K_\mathrm{b}$ of water is \(7\).

Step by step solution

01

Recall the relationship between pKw, pKa, and pKb#

pKw is the ionic product of water and it is equal to the sum of pKa and pKb. In other words, pKa + pKb = pKw. At 25°C, pKw has a value of 14.
02

Find pKa of water#

In water, the concentration of hydrogen ions (H+) is equal to the concentration of hydroxide ions (OH-). Since the pH of pure water is 7, we can find its pKa as follows: pKa = -log10[H+] pKa = -log10(10^{-7}) = 7
03

Determine the pKb of water#

Now that we have the pKa value, we can use the relationship between pKw, pKa, and pKb to find the pKb of water: pKb = pKw - pKa pKb = 14 - 7 pKb = 7
04

Choose the correct option#

From our calculation, the pKb of water is found to be 7. Therefore, the correct answer is (b) 7.

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Key Concepts

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

Ionic Product of Water
Diving into the intricate world of acids, bases, and their equilibrium in water can be a mind-bending journey. But let's simplify it! At the heart of this dynamic balance is the ionic product of water, denoted as Kw. This tiny constant is the result of water (\text{H}_2\text{O}) naturally dissociating into hydrogen ions (\text{H}^+) and hydroxide ions (\text{OH}^-), a process that's happening in every drop of water.

The ionic product of water is mathematically expressed as: \[Kw = [H^+][OH^-]\] At 25°C (room temperature), Kw is always 1.0 x 10^{-14} M^2, a small number reflecting the rarity of this event. Now, because the concentrations of \text{H}^+ and \text{OH}^- in pure water are equal, we can square root the Kw to find their individual concentrations, unearthing a value of 1.0 x 10^{-7} M for both. That's some neat chemistry magic right there!
pKa and pKb Relationship
When tackling the relationship between pKa and pKb, think of them as two sides of the same coin. These measures give us a sneak peek into the strengths of acids and bases. But how do they connect? Simply put:\
  • \text{pKa} is the negative logarithm of the acid dissociation constant \text{(Ka)}.
  • \text{pKb} is the negative logarithm of the base dissociation constant \text{(Kb)}.

The ionic product of water \text{(Kw)} ties them together, giving us the handy equation: \[pKw = pKa + pKb\] For water at 25°C, \text{pKw} is always 14. If you find the pKa of a substance, you just subtract it from 14 to get the corresponding pKb. Easy, right? For water, pKa equals pKb, making them each 7 since they're derived from the halving of \text{pKw}. That's the kind of symmetry that makes chemistry beautifully balanced.
pH and pOH Calculations
To make sense of how acidic or basic a solution is, we use pH and pOH scales—basically the bouncers that decide how strong or weak an acid or base is at the molecular party in a solution. Now, brace yourself for this: pH measures the concentration of hydrogen ions, while pOH does the same for hydroxide ions.

Calculating them is a walk in the park:
  • \text{pH} = -\text{log}([H^+])
  • \text{pOH} = -\text{log}([OH^-])

Remember, \text{pH} and \text{pOH} aren't independent freelancers; they work together to balance each other out. This duo always adds up to 14 in a solution at 25°C, summing up to the fixed value of \text{pKw}. So, when the pH dips like a limbo stick, the pOH jumps higher, and vice versa. Knowing one gives you a clear picture of the other: \[pH + pOH = 14\] Pretty neat, isn't it? Keeping track of both ensures your acid-base chemistry is always in harmony.

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Most popular questions from this chapter

Addition of the indicator methyl orange to an unknown solution leads to a yellow color. The addition of bromthymol blue to the same solution also leads to a yellow color. (a) Is the solution acidic, neutral, or basic? (b) What is the range (in whole numbers) of possible pH values for the solution? (c) Is there another indicator you could use to narrow the range of possible pH values for the solution?

(a) Given that \(K_{b}\) for ammonia is \(1.8 \times 10^{-5}\) and that for hydroxylamine is \(1.1 \times 10^{-8}\) , which is the stronger base? (b) Which is the stronger acid, the ammonium or the hydroxylammonium ion? (c) Calculate \(K_{a}\) values for \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{H}_{3} \mathrm{NOH}^{+}.\)

Label each of the following as being a strong base, a weak base, or a species with negligible basicity. In each case write the formula of its conjugate acid, and indicate whether the conjugate acid is a strong acid, a weak acid, or a species with negligible acidity: \((\mathbf{a})\mathrm{CH}_{3} \mathrm{COO}^{-},(\mathbf{b}) \mathrm{HCO}_{3}^{-},(\mathbf{c}) \mathrm{O}^{2-},(\mathbf{d}) \mathrm{Cl}^{-},(\mathbf{e}) \mathrm{NH}_{3}\)

Consider the base hydroxylamine, \(\mathrm{NH}_{2} \mathrm{OH}\) . (a) What is the conjugate acid of hydroxylamine? (b) When it acts as a base, which atom in hydroxylamine accepts a proton? (c) There are two atoms in hydroxylamine that have nonbonding electron pairs that could act as proton acceptors. Use Lewis structures and formal charges (Section 8.5) to rationalize why one of these two atoms is a much better proton acceptor than the other.

Indicate whether each of the following statements is correct or incorrect. (a) Every Bronsted-Lowry acid is also a Lewis acid. (b) Every Lewis acid is also a Bronsted-Lowry acid. (c) Conjugate acids of weak bases produce more acidic solutions than conjugate acids of strong bases. (d) \(\mathrm{K}^{+}\) ion is acidic in water because it causes hydrating water molecules to become more acidic. (e) The percent ionization of a weak acid in water increases as the concentration of acid decreases.

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