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A buffer solution is prepared by adding \(5.50 \mathrm{~g}\) of ammonium chloride and \(0.0188\) mol of ammonia to enough water to make \(155 \mathrm{~mL}\) of solution. (a) What is the \(\mathrm{pH}\) of the buffer? (b) If enough water is added to double the volume, what is the \(\mathrm{pH}\) of the solution?

Short Answer

Expert verified
Answer: (a) The pH of the buffer solution is 8.30. (b) The pH of the buffer solution after doubling the volume is 7.96.

Step by step solution

01

Find the concentrations of ammonium chloride and ammonia in the original solution.

To do this, we will find the moles of ammonium chloride and then divide it by the volume of the solution. Ammonium chloride: mass = 5.50 g Molar mass of NH4Cl = 53.49 g/mol moles of NH4Cl = mass / molar mass moles of NH4Cl = 5.50 g / 53.49 g/mol = 0.1028 mol Ammonia: moles of NH3 = 0.0188 mol Now, we will find the concentrations of each component in the solution. Volume of solution = 155 mL = 0.155 L Concentration of NH4Cl = moles of NH4Cl / volume of solution Concentration of NH4Cl = 0.1028 mol / 0.155 L = 0.663 M Concentration of NH3 = moles of NH3 / volume of solution Concentration of NH3 = 0.0188 mol / 0.155 L = 0.121 M
02

Use the Henderson-Hasselbalch equation to find the pH of the buffer solution.

The Henderson-Hasselbalch equation is: pH = pKa + log ([base]/[acid]) The reaction for this buffer solution is: NH3 + H2O <-> NH4+ + OH- pKa = -log(Ka) (for ammonia, NH3) The Ka value for ammonia is 1.8 × 10^(-5), so pKa = -log(1.8 × 10^(-5)) = 9.74 Now, we will find the pH: pH = 9.74 + log (0.121/0.663) pH = 9.74 - 1.44 = 8.30 The pH of the buffer solution is 8.30.
03

Find the new pH after diluting the buffer solution.

When we double the volume of the solution, the concentrations of NH4Cl and NH3 will be halved. New concentration of NH4Cl = 0.663 M / 2 = 0.3315 M New concentration of NH3 = 0.121 M / 2 = 0.0605 M Using the Henderson-Hasselbalch equation again: pH = 9.74 + log (0.0605/0.3315) pH = 9.74 - 1.78 = 7.96 The pH of the diluted buffer solution is 7.96. Answer: (a) The pH of the buffer solution is 8.30. (b) The pH of the buffer solution after doubling the volume is 7.96.

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Key Concepts

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

pH calculation
Understanding how to calculate the pH of a solution is crucial in chemistry. In this context, a buffer solution made from ammonium chloride and ammonia is at play. A buffer helps resist changes in pH when small amounts of acid or base are added.
When calculating the pH for a buffer solution, first determine the concentrations of the acidic (NH4Cl) and basic (NH3) components. Here's a recap of the calculation steps:
  • Calculate the moles of ammonium chloride using its mass and molar mass. This division gives the moles of NH4Cl.
  • Given moles need to be divided by the total solution volume (in liters) to find concentration, which applies to both NH4Cl and NH3.
  • Use the known concentration values in the Henderson-Hasselbalch equation to find the pH.
Remember, these calculations give the pH based on the buffer components present and their relative concentrations in the solution.
Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is a fundamental tool in chemistry, especially for dealing with buffer solutions. It's represented as:\[\text{pH} = \text{pK}_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)\]This equation allows you to calculate the pH of a buffer solution by incorporating the concentrations of the base and acid involved. In this exercise, ammonia functions as the base, and the ammonium ion derived from ammonium chloride serves as the acid.
The equation's core simplicity helps:
  • \( \text{pK}_a \) is the negative logarithm of the acid dissociation constant (\(K_a\)) of the weak acid component. For ammonia, it accounts for the species forming the buffer.
  • The division of the concentration of the base ([NH3]) by that of the acid ([NH4+]) gives the ratio required for the log term in the equation.
Manipulating this equation allows determination of how varying concentrations affect the pH of the solution, giving insight into the buffer's behavior under different conditions.
ammonium chloride
Ammonium chloride (NH4Cl) plays a crucial role in forming buffer solutions, particularly those involving ammonia (NH3). It's an ionic compound that readily dissolves in water, disassociating into ammonium (NH4+) and chloride (Cl-) ions.
The presence of NH4Cl in a buffer solution serves as the acidic component since NH4+ can release a proton (H+) to counter any added base and maintain pH stability. This feature is essential for creating a buffer solution with a specific pH range. Consider these key points:
  • Dissolution: NH4Cl dissolves completely in water, providing the necessary NH4+ ions for buffering reactions.
  • Equilibrium: The buffer conducts a dynamic equilibrium, working with NH3 to oppose pH changes even when dilute acid or base is introduced.
By understanding the mechanisms of ammonium chloride in buffer solutions, one can appreciate its pivotal role in stabilizing pH and sustaining equilibrium in chemical reactions.

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Most popular questions from this chapter

Consider the titration of \(\mathrm{HF}\left(K_{\mathrm{a}}=6.7 \times 10^{-4}\right)\) with \(\mathrm{NaOH}\). What is the \(\mathrm{pH}\) when a third of the acid has been neutralized?

A \(0.1375 \mathrm{M}\) solution of potassium hydroxide is used to titrate \(35.00 \mathrm{~mL}\) of \(0.257 M\) hydrobromic acid. (Assume that volumes are additive.) (a) Write a balanced net ionic equation for the reaction that takes place during titration. (b) What are the species present at the equivalence point? (c) What volume of potassium hydroxide is required to reach the equivalence point? (d) What is the \(\mathrm{pH}\) of the solution before any \(\mathrm{KOH}\) is added? (e) What is the \(\mathrm{pH}\) of the solution halfway to the equivalence point? (f) What is the \(\mathrm{pH}\) of the solution at the equivalence point?

Given three acid-base indicators-methyl orange (end point at \(\mathrm{pH}\) 4), bromthymol blue (end point at \(\mathrm{pH} 7\) ), and phenolphthalein (end point at \(\mathrm{pH}\) 9) - which would you select for the following acid-base titrations? (a) perchloric acid with an aqueous solution of ammonia (b) nitrous acid with lithium hydroxide (c) hydrobromic acid with strontium hydroxide (d) sodium fluoride with nitric acid

Which of the following would form a buffer if added to \(650.0 \mathrm{~mL}\) of \(0.40 M \mathrm{Sr}(\mathrm{OH})_{2} ?\) (a) \(1.00 \mathrm{~mol}\) of \(\mathrm{HF}\) (b) \(0.75 \mathrm{~mol}\) of \(\mathrm{HF}\) (c) \(0.30 \mathrm{~mol}\) of \(\mathrm{HF}\) (d) \(0.30 \mathrm{~mol}\) of \(\mathrm{NaP}\) (e) \(0.30 \mathrm{~mol}\) of \(\mathrm{HCl}\) Explain your reasoning in each case.

A buffer is made up of \(0.300\) L each of \(0.500 \mathrm{MKH}_{2} \mathrm{PO}_{4}\) and \(0.317 \mathrm{M}\) \(\mathrm{K}_{2} \mathrm{HPO}_{4}\). Assuming that volumes are additive, calculate (a) the \(\mathrm{pH}\) of the buffer. (b) the \(\mathrm{pH}\) of the buffer after the addition of \(0.0500 \mathrm{~mol}\) of \(\mathrm{HCl}\) to \(0.600 \mathrm{~L}\) of buffer. (c) the \(\mathrm{pH}\) of the buffer after the addition of \(0.0500 \mathrm{~mol}\) of \(\mathrm{NaOH}\) to \(0.600 \mathrm{~L}\) of buffer.

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