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Chapter 19: Problem 25

What is the component concentration ratio, \(\left[\mathrm{NO}_{2}^{-}\right] /\left[\mathrm{HNO}_{2}\right],\) of a buffer that has a \(\mathrm{pH}\) of \(2.95\left(K_{\mathrm{a}}\right.\) of \(\left.\mathrm{HNO}_{2}=7.1 \times 10^{-4}\right) ?\)

Short Answer

Expert verified
\frac{[\text{NO}_2^-]}{[\text{HNO}_2]} = 0.63

Step by step solution

01

Identify the Necessary Equation

Use the Henderson-Hasselbalch equation to relate \(\text{pH}\) to the component concentration ratio. The equation is \(\text{pH} = \text{pKa} + \text{log} \frac{[\text{A}^-]}{[\text{HA}]}\), where \( \text{A}^- \) is the concentration of the conjugate base and \( \text{HA} \) is the concentration of the weak acid.
02

Calculate \( \text{pKa} \)

Use the given \( K_a \) of \( \text{HNO}_2 \) to find \(\text{pKa}\). The formula to convert \( K_a \) to \( \text{pKa} \) is \(\text{pKa} = -\text{log}(K_a)\). Given \( K_a = 7.1 \times 10^{-4} \), calculate \( \text{pKa} = -\text{log}(7.1 \times 10^{-4}) \).
03

Perform the Calculation

Perform the calculation: \( \text{pKa} = -\text{log}(7.1 \times 10^{-4}) \). This will give \(\text{pKa} = 3.15\).
04

Substitute Values into Henderson-Hasselbalch Equation

Substitute the given \( \text{pH} \) and the calculated \( \text{pKa} \) into the Henderson-Hasselbalch equation: \( 2.95 = 3.15 + \text{log} \frac{[\text{NO}_2^-]}{[\text{HNO}_2]} \).
05

Solve for the Component Concentration Ratio

Isolate the log term to solve for the component concentration ratio: \(\text{log} \frac{[\text{NO}_2^-]}{[\text{HNO}_2]} = 2.95 - 3.15\), which simplifies to \( \text{log} \frac{[\text{NO}_2^-]}{[\text{HNO}_2]} = -0.20 \). Use the inverse log to find the ratio: \(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]} = 10^{-0.20}\).
06

Calculate the Final Ratio

Perform the final calculation: \(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]} = 10^{-0.20} = 0.63 \).

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Key Concepts

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

Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is an essential equation in chemistry that helps us calculate the pH of buffer solutions. It connects the pH of a solution with the concentrations of a weak acid and its conjugate base. The equation is:\[ \text{pH} = \text{pKa} + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \]In this equation, \( [A^-] \) represents the concentration of the conjugate base, and \[ [HA] \] represents the concentration of the weak acid.
By rearranging and substituting values into this equation, students can easily find the ratio between the conjugate base and the weak acid in a buffer solution.
Weak acid
Weak acids are acids that do not fully dissociate in water. Instead, they partially dissociate, reaching an equilibrium between the undissociated acid (HA) and the ions produced (H^+) and (A^-).
Examples of weak acids include acetic acid (vinegar) and nitrous acid (HNO2).
Their partial dissociation is what makes buffer solutions effective in resisting pH changes.
Conjugate base
A conjugate base is what remains after a weak acid has donated a proton (H+). For example, in the dissociation of nitrous acid (HNO2), the conjugate base is the nitrite ion (NO2-).
The strength and concentration of the conjugate base, in combination with the weak acid, determine the effectiveness of a buffer solution.
Understanding the relationship between the conjugate base and its weak acid counterpart is crucial when using the Henderson-Hasselbalch equation.
pKa
The pKa of a weak acid is a crucial figure that indicates its acid strength. It is derived from the acid dissociation constant (Ka).
Mathematically, pKa is represented as:\[ \text{pKa} = -\text{log}(K_a) \]A lower pKa value indicates a stronger weak acid, whereas a higher one signifies a weaker acid.
Knowing the pKa allows us to use the Henderson-Hasselbalch equation effectively to determine the pH or component concentration ratios like we did in the given exercise.
Ka
Ka, or the acid dissociation constant, quantifies the extent to which an acid dissociates in water. It is a measure of the ionization of a weak acid. The Ka expression for a weak acid HA in water is:\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]A larger Ka means a stronger acid, indicating more dissociation into ions.
For instance, in our exercise, we needed the Ka of HNO2 to determine its pKa, ultimately helping us find the solution's pH and component concentration ratio.
Understanding Ka is fundamental when discussing weak acids and buffer solutions.

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Most popular questions from this chapter

Which compound in each pair is more soluble in water? (a) Strontium sulfate or barium chromate (b) Calcium carbonate or copper(II) carbonate (c) Barium iodate or silver chromate

Write equations to show whether the solubility of either of the following is affected by \(\mathrm{pH}\) : (a) \(\mathrm{AgCl} ;\) (b) \(\mathrm{SrCO}_{3}\).

Ethylenediaminetetraacetic acid (abbreviated \(\mathrm{H}_{4} \mathrm{EDTA}\) ) is a tetraprotic acid. Its salts are used to treat toxic metal poisoning by forming soluble complex ions that are then excreted. Because EDTA \(^{4-}\) also binds essential calcium ions, it is often administered as the calcium disodium salt. For example, when \(\mathrm{Na}_{2} \mathrm{Ca}\) (EDTA) is given to a patient, the \([\mathrm{Ca}(\mathrm{EDTA})]^{2-}\) ions react with circulating \(\mathrm{Pb}^{2+}\) ions and the metal ions are exchanged: \([\mathrm{Ca}(\mathrm{EDTA})]^{2-}(a q)+\mathrm{Pb}^{2+}(a q) \rightleftharpoons\) $$ [\mathrm{Pb}(\mathrm{EDTA})]^{2-}(a q)+\mathrm{Ca}^{2+}(a q) \quad K_{\mathrm{c}}=2.5 \times 10^{7} $$ A child has a dangerous blood lead level of \(120 \mu \mathrm{g} / 100 \mathrm{~mL}\). If the child is administered \(100 . \mathrm{mL}\) of \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{Ca}(\mathrm{EDTA}),\) what is the final concentration of \(\mathrm{Pb}^{2+}\) in \(\mu \mathrm{g} / 100 \mathrm{~mL}\) blood, assuming that the exchange reaction and excretion process are \(100 \%\) efficient? (Total blood volume is \(1.5 \mathrm{~L} .)\)

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 \(\left.\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}=4.7 \times 10^{13}\right]\)

Find the \(\mathrm{pH}\) of a buffer that consists of \(0.45 \mathrm{M} \mathrm{HCOOH}\) and \(0.63 M \mathrm{HCOONa}\left(\mathrm{p} K_{\mathrm{a}}\right.\) of \(\left.\mathrm{HCOOH}=3.74\right)\)
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