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Chapter 19: Problem 28

A buffer containing \(0.2000 M\) of acid, HA, and \(0.1500 M\) of its conjugate base, \(\mathrm{A}^{-},\) has a \(\mathrm{pH}\) of \(3.35 .\) What is the \(\mathrm{pH}\) after \(0.0015 \mathrm{~mol}\) of \(\mathrm{NaOH}\) is added to \(0.5000 \mathrm{~L}\) of this solution?

Short Answer

Expert verified
The new pH is 3.37

Step by step solution

01

- Identify initial concentrations

Given concentrations: HA = 0.2000 MA⁻ = 0.1500 MpH = 3.35
02

- Determine initial moles in 0.5000 L

Initial moles of HA: 0.2000 M * 0.5000 L = 0.1000 molInitial moles of A⁻: 0.1500 M * 0.5000 L = 0.0750 mol
03

- Calculate moles of HA and A⁻ after NaOH addition

0.0015 mol of NaOH will react with HA: Moles of HA after reaction: 0.1000 mol - 0.0015 mol = 0.0985 molMoles of A⁻ after reaction: 0.0750 mol + 0.0015 mol = 0.0765 mol
04

- Calculate new concentrations

New concentration of HA in 0.5000 L:0.0985 mol / 0.5000 L = 0.1970 MNew concentration of A⁻ in 0.5000 L: 0.0765 mol / 0.5000 L = 0.1530 M
05

- Use the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is \( \text{pH} = \text{p}K_\text{a} + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \)Calculate \text{pK}_\text{a}: \( \text{pH} = \text{p}K_\text{a} + \text{log} \frac{0.1500}{0.2000} \)\( 3.35 = \text{p}K_\text{a} - 0.1249 \)\( \text{p}K_\text{a} = 3.4749 \)
06

- Calculate the new pH

Using the new concentrations:\( \text{pH} = \text{p}K_\text{a} + \text{log} \frac{0.1530}{0.1970} \)\( \text{pH} = 3.4749 + \text{log} (0.7766) \)\( \text{pH} = 3.4749 - 0.1095 \)\( \text{pH} = 3.3654 \)

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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 essential for calculating the pH of buffer solutions. It relates the pH, the dissociation constant (\text{p}K_\text{a}), and the concentrations of an acid and its conjugate base.

This equation is written as: \( \text{pH} = \text{p}K_\text{a} + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \)
where:
  • \(\text{pH}\) is the measure of acidity
  • \(\text{p}K_\text{a}\) is the negative logarithm of the acid dissociation constant
  • \([\text{A}^-]\) is the concentration of the conjugate base
  • \([\text{HA}]\) is the concentration of the acid
The formula allows you to understand how changes in the concentrations of the conjugate base or acid affect the pH of the solution.

In the given problem, after determining the initial and new concentrations of HA and A⁻, the Henderson-Hasselbalch equation helps us find the new pH after adding a small amount of NaOH to the buffer solution.
Buffer capacity
Buffer capacity refers to the ability of a buffer solution to resist changes in pH when an acid or base is added. It depends on two main factors:
  • The concentrations of the acid and its conjugate base
  • The absolute amounts of HA and A⁻ in the solution
A buffer is more effective when the concentrations of the acid and its conjugate base are high and nearly equal.

In our exercise, the buffer includes 0.200 M of HA and 0.150 M of A⁻. The buffer capacity is demonstrated when 0.0015 mol of NaOH (a base) is added.
Due to the buffer’s composition, the pH of the solution changes only slightly from 3.35 to 3.3654, showing its resistance to pH alterations.
Acid-base equilibrium
Acid-base equilibrium involves the balance between acids and bases in a solution. When an acid (HA) donates a proton (H\textsuperscript{+}), it forms its conjugate base (A⁻).

The equilibrium expression is written as:
\[\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\]
This balance is crucial for the buffer's ability to maintain pH.

In our exercise, when NaOH is added, it reacts with HA (acid) to form water and A⁻ (conjugate base):
  • \text{NaOH} + \text{HA} \rightarrow \text{H}_2\text{O} + \text{A}^-
By this reaction, HA decreases and A⁻ increases, but the equilibrium adjusts so that pH only changes slightly.
Understanding this equilibrium helps explain why the buffer resists significant changes in pH.
Concentration calculations
Concentration calculations are fundamental in chemistry to determine the amount of substances in a given volume.

For the exercise, we first calculate the initial moles of HA and A⁻ in 0.5000 L of the solution:
  • Initial moles of HA: \( 0.2000 \text{ M} \times 0.5000 \text{ L} = 0.1000 \text{ mol} \)
  • Initial moles of A⁻: \( 0.1500 \text{ M} \times 0.5000 \text{ L} = 0.0750 \text{ mol} \)
After adding 0.0015 mol of NaOH, the moles change because HA reacts with NaOH:
  • New moles of HA: \( 0.1000 \text{ mol} - 0.0015 \text{ mol} = 0.0985 \text{ mol} \)
  • New moles of A⁻: \( 0.0750 \text{ mol} + 0.0015 \text{ mol} = 0.0765 \text{ mol} \)
Next, we determine the new concentrations by dividing the new moles by the volume of the solution (0.5000 L):
  • New concentration of HA: \( \frac{0.0985 \text{ mol}}{0.5000 \text{ L}} = 0.1970 \text{ M} \)
  • New concentration of A⁻: \( \frac{0.0765 \text{ mol}}{0.5000 \text{ L}} = 0.1530 \text{ M} \)
These calculations are vital for applying the Henderson-Hasselbalch equation and understanding the buffer's response to the added base.

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Most popular questions from this chapter

The solubility of \(\mathrm{Ag}(\mathrm{I})\) in aqueous solutions containing different concentrations of \(\mathrm{Cl}^{-}\) is based on the following equilibria: $$ \begin{array}{lr} \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) & \rightleftharpoons \mathrm{AgCl}(s) & K_{\mathrm{sp}}=1.8 \times 10^{-10} \\ \mathrm{Ag}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) & \rightleftharpoons \mathrm{AgCl}_{2}^{-}(a q) & K_{\mathrm{f}}=1.8 \times 10^{5} \end{array} $$ When solid \(\mathrm{AgCl}\) is shaken with a solution containing \(\mathrm{Cl}^{-}, \mathrm{Ag}(\mathrm{I})\) is present as both \(\mathrm{Ag}^{+}\) and \(\mathrm{AgCl}_{2}^{-}\). The solubility of \(\mathrm{AgCl}\) is the sum of the concentrations of \(\mathrm{Ag}^{+}\) and \(\mathrm{AgCl}_{2}^{-}\) (a) Show that \(\left[\mathrm{Ag}^{+}\right]\) in solution is given by $$ \left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-10} /\left[\mathrm{Cl}^{-}\right] $$ and that \(\left[\mathrm{AgCl}_{2}^{-}\right]\) in solution is given by $$ \left[\mathrm{AgCl}_{2}^{-}\right]=\left(3.2 \times 10^{-5}\right)\left(\left[\mathrm{Cl}^{-}\right]\right) $$ (b) Find the \(\left[\mathrm{Cl}^{-}\right]\) at which \(\left[\mathrm{Ag}^{+}\right]=\left[\mathrm{AgCl}_{2}^{-}\right]\). (c) Explain the shape of a plot of \(\mathrm{AgCl}\) solubility vs. \(\left[\mathrm{Cl}^{-}\right]\). (d) Find the solubility of \(\mathrm{AgCl}\) at the \(\left[\mathrm{Cl}^{-}\right]\) of part \((\mathrm{b}),\) which is the minimum solubility of \(\mathrm{AgCl}\) in the presence of \(\mathrm{Cl}^{-}\).

What is the difference between the end point of a titration and the equivalence point? Is the equivalence point always reached first? Explain.

Find the solubility of \(\mathrm{Cr}(\mathrm{OH})_{3}\) in a buffer of \(\mathrm{pH} 13.0\left[K_{\mathrm{sp}}\right.\) of \(\mathrm{Cr}(\mathrm{OH})_{3}=6.3 \times 10^{-31} ; K_{\mathrm{f}}\) of \(\left.\mathrm{Cr}(\mathrm{OH})_{4}^{-}=8.0 \times 10^{29}\right]\)

The normal \(\mathrm{pH}\) of blood is \(7.40 \pm 0.05\) and is controlled in part by the \(\mathrm{H}_{2} \mathrm{CO}_{3} / \mathrm{HCO}_{3}^{-}\) buffer system. (a) Assuming that the \(K_{\mathrm{a}}\) value for carbonic acid at \(25^{\circ} \mathrm{C}\) applies to blood, what is the \(\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right] /\left[\mathrm{HCO}_{3}^{-}\right]\) ratio in normal blood? (b) In a condition called acidosis, the blood is too acidic. What is the \(\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right] /\left[\mathrm{HCO}_{3}^{-}\right]\) ratio in a patient whose blood \(\mathrm{pH}\) is \(7.20 ?\)

Write the ion-product expressions for (a) iron(III) hydroxide; (b) barium phosphate; (c) magnesium fluoride.
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