Examples of ka in the following topics:
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- The Ka expression is as follows:
- The logarithmic constant (pKa) is equal to -log10(Ka).
- The larger the value of pKa, the smaller the extent of dissociation.
- Acids with a pKa value of less than about -2 are said to be strong acids.
- What is the pKa for acetic acid?
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- Weak acids have very small values for Ka (and therefore higher values for pKa) compared to strong acids, which have very large Ka values (and slightly negative pKa values).
- The Ka of weak acids varies between 1.8×10−16 and 55.5.
- The first Ka refers to the first dissociation step:
- This Ka value is 4.46×10−7 (pKa1 = 6.351).
- The Ka of acetic acid is $1.8\times 10^{-5}$.
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- Ka is a quantitative measure of the strength of an acid in solution.
- The logarithmic constant, pKa, which is equal to −log10 (Ka), is sometimes incorrectly referred to as an acid dissociation constant as well.
- Smaller Ka values yield larger pKa values.
- An understanding of Ka is also essential for working with buffers; the design of these solutions depends on a knowledge of the pKa values of their components.
- Another important application of Ka is with pH indicators.
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- More precisely, the acid must be stronger in aqueous solution than a hydronium ion (H+), so strong acids have a pKa < -1.74.
- An example is hydrochloric acid (HCl), whose pKa is -6.3.
- p-Toluenesulfonic acid is an example of an organic soluble strong acid, with a pKa of -2.8.
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- The Henderson–Hasselbalch equation connects the measurable value of the pH of a solution with the theoretical value pKa.
- The Henderson–Hasselbalch equation mathematically connects the measurable pH of a solution with the pKa (which is equal to -log Ka) of the acid.
- The equation can be derived from the formula of pKa for a weak acid or buffer.
- With a given pH and known pKa, the solution of the Henderson-Hasselbalch equation gives the logarithm of a ratio which can be solved by performing the antilogarithm of pH/pKa:
- What is the pH of a buffer solution consisting of 0.0350 M NH3 and 0.0500 M NH4+ (Ka for NH4+ is 5.6 x 10-10)?
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- Percent dissociation represents an acid's strength and can be calculated using the Ka value and the solution's pH.
- We have already discussed quantifying the strength of a weak acid by relating it to its acid equilibrium constant Ka; now we will do so in terms of the acid's percent dissociation.
- Calculate the percent dissociation of a weak acid in a $0.060\;M$ solution of HA ($K_a=1.5\times 10^{-5}$).
- Calculate percent dissociation for weak acids from their Ka values and a given concentration.
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- The exact ratio of the base to the acid for a desired pH can be determined from the Ka value and the Henderson-Hasselbalch equation.
- Of the acids listed, the Ka value for acetic acid is closest to the desired hydrogen ion concentration.
- The pKa of acetic acid is
- Extrapolating further from this, a buffer is most effective when the concentrations of acid and conjugate base (or base and conjugate acid) are approximately equal—in other words, when the log [base]/[acid] equals 0 and the pH equals the pKa.
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- These acids can be arranged in order of their pKa values and, by extension, their relative strengths:
- HOCl pKa = 7.5 < HOBr pKa = 8.6 < HOI pKa = 10.6
- Recall that smaller values of pKa correspond to greater acid strength.
- Consider the family of chlorooxoacids, which are arranged below in order of pKa values:
- HOClO3 pKa = -8 < HOClO2 pKa = -1.0 < HOClO pKa = 1.92 < HOCl pKa = 7.53
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- An acid dissociation constant, Ka, is the equilibrium constant for the dissociation of an acid in aqueous solution.
- As such, strong acids will have large values of Ka that are greater than one, which indicates that the forward reaction of dissociation is strongly favored.
- Weak acids, on the other hand, will have small values of Ka that are less than one, indicating that the reverse reaction is strongly favored; weak acids dissociate only to a small extent.
- As such, Ka acts a relative indicator of acid strength.
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- We can determine the answer by comparing Ka and Kb values for each ion.
- In this case, the value of Kb for bicarbonate is greater than the value of Ka for ammonium.
- In summary, when a salt contains two ions that hydrolyze, compare their Ka and Kb values: