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Chapter 19: Problem 26

What is the component concentration ratio, \(\left[\mathrm{BrO}^{-}\right] /[\mathrm{HBrO}]\), of a buffer that has a \(\mathrm{pH}\) of \(7.95\left(K_{\mathrm{a}}\right.\) of \(\left.\mathrm{HBrO}=2.3 \times 10^{-9}\right) ?\)

Short Answer

Expert verified
[BrO⁻]/[HBrO] = 0.204

Step by step solution

01

Write the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentration of the acid and its conjugate base. The equation is: \[ \text{pH} = \text{p}K_a + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \]
02

Calculate the pKa from the given Ka

The pKa is the negative logarithm of the Ka value. Use the following formula: \[ \text{p}K_a = -\text{log}(K_a) \]Given that \[ K_a = 2.3 \times 10^{-9} \]Calculate \[ \text{p}K_a = -\text{log}(2.3 \times 10^{-9}) \ = 8.64 \]
03

Insert values into the Henderson-Hasselbalch equation

Substitute the known values into the equation: \[ 7.95 = 8.64 + \text{log} \frac{[\text{BrO}^-]}{[\text{HBrO}]} \]
04

Solve for the ratio \([\text{BrO}^-] / [\text{HBrO}]\)

Rearrange the equation to isolate the log term: \[ 7.95 - 8.64 = \text{log} \frac{[\text{BrO}^-]}{[\text{HBrO}]} \]Calculate the difference: \[ -0.69 = \text{log} \frac{[\text{BrO}^-]}{[\text{HBrO}]} \]
05

Eliminate the log by exponentiating both sides

To find the ratio, exponentiate both sides: \[ 10^{-0.69} = \frac{[\text{BrO}^-]}{[\text{HBrO}]} \]Calculate the value: \[ 10^{-0.69} = 0.204 \]Thus, the ratio \[ \frac{[\text{BrO}^-]}{[\text{HBrO}]} = 0.204 \]

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Key Concepts

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

buffer solution
A buffer solution is a mixture of a weak acid and its conjugate base, or a weak base and its conjugate acid. This solution resists changes in pH when small amounts of an acid or base are added to it.
Buffers are crucial in many biological and chemical systems where a stable pH is necessary.
For example, blood is a natural buffer that maintains a pH around 7.4 to ensure proper physiological function.
pH calculation
The pH of a solution measures its acidity or alkalinity. It is calculated using the formula:
\[ \text{pH} = -\text{log}[\text{H}^+] \]
For acids and bases, pH values range from 0 to 14. A pH of 7 is neutral, values below 7 are acidic, and values above 7 are basic.
The Henderson-Hasselbalch equation is a useful tool for calculating the pH of a buffer solution. It relates the pH to the pKa and the ratio of the concentrations of the conjugate base and acid:\br>
\[ \text{pH} = \text{p}K_a + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \].
component concentration ratio
The component concentration ratio is the ratio of the concentrations of the conjugate base to the weak acid in a buffer solution ([A-]/[HA]).
This ratio is essential for determining the pH of the buffer using the Henderson-Hasselbalch equation.
In our example, the component concentration ratio is calculated as follows: \[ \frac{[\text{BrO}^-]}{[\text{HBrO}]} = 0.204 \] A ratio less than 1 indicates a higher concentration of the acid than the conjugate base, which means the buffer will be more acidic.
acid dissociation constant
The acid dissociation constant (\text{K\textsubscript{a}}) is a measure of the strength of an acid in solution. It is the equilibrium constant for the dissociation of a weak acid into its conjugate base and a proton.
Mathematically, it is expressed as:
\[{K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}}\]
In our example, \textsubscript{HBrO} has a \textsubscript{K\textsubscript{a}}\text{ = 2.3 \times 10^{-9}}, indicating that it is a weak acid, as it dissociates very little in solution.
pKa calculation
The \textsubscript{pK\textsubscript{a}}\text{} is the negative logarithm of the acid dissociation constant (\textsubscript{K\textsubscript{a}}):
\[ \text{p}K\textsubscript{\textsubscript{a}} = -\text{log}(\text{K\textsubscript{a}}) \]
The \textsubscript{pK\textsubscript{a}}\text{} provides an easy way to express the strength of an acid. A lower \textsubscript{pK\textsubscript{a}}\text{} value indicates a stronger acid, as it dissociates more in solution.
For example, if \textsubscript{K\textsubscript{a}} is 2.3 \times 10^{-9}, the \textsubscript{pK\textsubscript{a}}\text{} is: \br> \[-\text{log}(2.3 \times 10^{-9}) = 8.64\]

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Most popular questions from this chapter

What is \(\left[\mathrm{Ag}^{+}\right]\) when \(25.0 \mathrm{~mL}\) each of \(0.044 \mathrm{M} \mathrm{AgNO}_{3}\) and \(0.57 M \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) are mixed? \(\left[K_{\mathrm{f}}\right.\) of \(\left.\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}=4.7 \times 10^{13}\right]\)

A buffer that contains \(0.110 M \mathrm{HY}\) and \(0.220 \mathrm{M} \mathrm{Y}^{-}\) has a \(\mathrm{pH}\) of 8.77 . What is the \(\mathrm{pH}\) after \(0.0015 \mathrm{~mol}\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) is added to \(0.350 \mathrm{~L}\) of this solution?

The solubility of silver dichromate at \(15^{\circ} \mathrm{C}\) is \(8.3 \times 10^{-3} \mathrm{~g} / 100 . \mathrm{mL}\) solution. Calculate its \(K_{\mathrm{sp}}\)

An ecobotanist separates the components of a tropical bark extract by chromatography. She discovers a large proportion of quinidine, a dextrorotatory isomer of quinine used for control of arrhythmic heartbeat. Quinidine has two basic nitrogens \(\left(K_{\mathrm{b} 1}=4.0 \times 10^{-6}\right.\) and \(\left.K_{\mathrm{b} 2}=1.0 \times 10^{-\mathrm{i} 0}\right) .\) To measure the concentration of quinidine, she carries out a titration. Because of the low solubility of quinidine, she first protonates both nitrogens with excess \(\mathrm{HCl}\) and titrates the acidified solution with standardized base. A 33.85-mg sample of quinidine \((\mathscr{M}=324.41 \mathrm{~g} / \mathrm{mol})\) is acidified with \(6.55 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{HCl}\) (a) How many milliliters of \(0.0133 \mathrm{M} \mathrm{NaOH}\) are needed to titrate the excess HCl? (b) How many additional milliliters of titrant are needed to reach the first equivalence point of quinidine dihydrochloride? (c) What is the \(\mathrm{pH}\) at the first equivalence point?

Human blood contains one buffer system based on phosphate species and one based on carbonate species. Assuming that blood has a normal \(\mathrm{pH}\) of \(7.4,\) what are the principal phosphate and carbonate species present? What is the ratio of the two phosphate species? (In the presence of the dissolved ions and other species in blood, \(K_{\mathrm{a} 1}\) of \(\mathrm{H}_{3} \mathrm{PO}_{4}=1.3 \times 10^{-2}, K_{\mathrm{a} 2}=2.3 \times 10^{-7},\) and \(K_{\mathrm{a} 3}=6 \times 10^{-12} ; K_{\mathrm{a} 1}\) of \(\mathrm{H}_{2} \mathrm{CO}_{3}=8 \times 10^{-7}\) and \(\left.K_{\mathrm{a} 2}=1.6 \times 10^{-10} .\right)\)
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