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Chapter 17: Problem 8

What concentration of ammonia, \(\left[\mathrm{NH}_{3}\right],\) should be present in a solution with \(\left[\mathrm{NH}_{4}^{+}\right]=0.732 \mathrm{M}\) to produce a buffer solution with \(\mathrm{pH}=9.12 ?\) For \(\mathrm{NH}_{3}\) \(K_{\mathrm{h}}=1.8 \times 10^{-5}\)

Short Answer

Expert verified
The concentration of ammonia in the solution should be 240.45 M.

Step by step solution

01

Understand the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is: \( pH = pK_a + log \left(\frac{[Base]}{[Acid]}\right)\) Here, we'll modify this a bit for the given exercise where \(K_h = [H_3O^+][NH_3]/[NH_4^+]\). Taking negative logarithm on both sides we get \(pH = pK_a + log \left(\frac{[NH_3]}{[NH_4^+]}\right)\)
02

Calculate the \(pK_a\)

The \(pK_a\) is calculated from the \(K_h\) by taking the negative logarithm of \(K_h\): \(pK_a = -log(K_h)\); given \(K_h = 1.8 \times 10^{-5}\), hence \(pK_a = -log(1.8 \times 10^{-5}) = 4.74\)
03

Solve the Henderson Hasselbalch equation for the concentration of ammonia

Now that we have \(pH\), \(pK_a\) and \([NH_4^+]\), we can solve for \([NH_3]\) in the Henderson-Hasselbalch equation: \(9.12 = 4.74 + log \left(\frac{[NH_3]}{0.732}\right)\); Solving this gives the ammonia concentration: \([NH_3] = 0.732 \times 10^{(9.12 - 4.74)} = 0.732 \times 10^{4.38} = 240.45 \,M \)

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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 a simple yet powerful tool for understanding the relationship between the pH of a solution and the concentrations of its acid and base components. It's mainly used for buffer solutions, where we aim to maintain a stable pH. The equation can be expressed as: \[ pH = pK_a + \log \left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]In the context of our problem, ammonia (\(\text{NH}_3\)) serves as the base, and ammonium (\(\text{NH}_4^+\)) is the acid. This equation allows us to connect the pH of the buffer to the concentrations of \([\text{NH}_3]\) and \([\text{NH}_4^+]\). By adjusting the ratio of these components, we can achieve a desired pH.
Ammonia Concentration
Understanding ammonia concentration is crucial, especially when preparing buffer solutions. In our example, we're asked to find \([\text{NH}_3]\) that will give a specific pH. The given concentration of ammonium is 0.732 M. This means the challenge is to calculate the right amount of ammonia to add.To solve this, we rearrange the Henderson-Hasselbalch equation:- We know the target pH is 9.12.- We鈥檝e calculated \(pK_a\) as 4.74 from the given \(K_h\).Using these values, we solve for the ammonia concentration through:\[ [\text{NH}_3] = [\text{NH}_4^+] \times 10^{(pH - pK_a)} \]This results in an ammonia concentration of 240.45 M, which ensures the buffer solution maintains the specified pH of 9.12.
pH Calculation
pH is a metric used to express the acidity or basicity of a solution. For buffer systems, predicting and controlling the pH is essential. The pH scale ranges from 0 to 14, with values below 7 being acidic, and above 7 being basic.For our exercise, the desired pH is a specific value of 9.12, indicating a slightly basic solution. By employing the Henderson-Hasselbalch equation, we combine theoretical calculations with practical needs to achieve this exact pH.
Here's how it works:
  • Determine the \(pK_a\) of the acid-base pair. In our case, \(pK_a = 4.74\).
  • Use the concentrations of acid (\([\text{NH}_4^+]\)) and base (\([\text{NH}_3]\)) to find their ratio.
  • Insert these values into the equation to confirm the pH is properly balanced.
This process highlights the elegance of how seemingly complex chemical properties can be understood and applied with just a few calculations.

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Most popular questions from this chapter

Phenol red indicator changes from yellow to red in the pH range from 6.6 to \(8.0 .\) Without making detailed calculations, state what color the indicator will assume in each of the following solutions: (a) \(0.10 \mathrm{M} \mathrm{KOH}\) (b) \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} ;\) (c) \(0.10 \mathrm{M} \mathrm{NH}_{4} \mathrm{NO}_{3} ;\) (d) \(0.10 \mathrm{M}\) HBr; (e) \(0.10 \mathrm{M} \mathrm{NaCN} ;\) (f) \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}-0.10 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO}\).

Calculate the pH of the buffer formed by mixing equal volumes \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\right]=1.49 \mathrm{M} \quad\) with \(\quad\left[\mathrm{HClO}_{4}\right]=\) 1.001 M. \(K_{\mathrm{b}}=4.3 \times 10^{-4}\)

The effect of adding \(0.001 \mathrm{mol} \mathrm{KOH}\) to 1.00 Lof a solution that is \(0.10 \mathrm{M} \mathrm{NH}_{3}-0.10 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) is to (a) raise the pH very slightly; (b) lower the pH very slightly; (c) raise the pH by several units; (d) lower the pH by several units.

Carbonic acid is a weak diprotic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\) with \(K_{a_{1}}=4.43 \times 10^{-7}\) and \(K_{\mathrm{a}_{2}}=4.73 \times 10^{-11} .\) The equiv- alence points for the titration come at approximately pH 4 and 9. Suitable indicators for use in titrating carbonic acid or carbonate solutions are methyl orange and phenolphthalein. (a) Sketch the titration curve that would be obtained in titrating a sample of \(\mathrm{NaHCO}_{3}(\mathrm{aq})\) with \(1.00 \mathrm{M} \mathrm{HCl}\) (b) Sketch the titration curve for \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) with 1.00 M HCl. (c) What volume of \(0.100 \mathrm{M} \mathrm{HCl}\) is required for the complete neutralization of \(1.00 \mathrm{g} \mathrm{NaHCO}_{3}(\mathrm{s}) ?\) (d) What volume of \(0.100 \mathrm{M} \mathrm{HCl}\) is required for the complete neutralization of \(1.00 \mathrm{g} \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s}) ?\) (e) A sample of NaOH contains a small amount of \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) For titration to the phenolphthalein end point, \(0.1000 \mathrm{g}\) of this sample requires \(23.98 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{HCl} .\) An additional \(0.78 \mathrm{mL}\) is required to reach the methyl orange end point. What is the percent \(\mathrm{Na}_{2} \mathrm{CO}_{3},\) by mass, in the sample?

Even though the carbonic acid-hydrogen carbonate buffer system is crucial to the maintenance of the \(\mathrm{pH}\) of blood, it has no practical use as a laboratory buffer solution. Can you think of a reason(s) for this?
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