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You are asked to prepare a buffer solution with a pH of 3.50. The following solutions, all \(0.100 \mathrm{M},\) are available to you: HCOOH, CH \(_{3} \mathrm{COOH}, \mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{NaCHOO}\) \(\mathrm{NaCH}_{3} \mathrm{COO},\) and \(\mathrm{NaH}_{2} \mathrm{PO}_{4} . \quad\) Describe how you would prepare this buffer solution. [Hint: What volumes of which solutions would you use?

Short Answer

Expert verified
To make a buffer solution of pH 3.50, we should mix the formic acid HCOOH and sodium formate NaHCOO solutions, taking 1 liter of formic acid for every 0.58 liters of sodium formate solution.

Step by step solution

01

Choosing the right solution to create a buffer

From the given \(0.100 \mathrm{M}\) solutions, the correct one to choose would be the one that yields a buffer with a pH of 3.50. We choose the weak acid and its conjugate base fron the list provided: HCOOH (Formic acid) and NaCHO (\( \text {Sodium formate}\))
02

Calculate the dissociation constant for Formic acid

The first step in preparing a buffer solution is to find the dissociation constant, pKa, of the chosen weak acid from a standard chemistry data tables. pKa of a substance is related to pH and indicates how readily hydrogen ions are given off by the acid in the solution. For Formic acid the pKa value is 3.74.
03

Applying the Henderson-Hasselbalch equation

For preparing a buffer solution, the Henderson-Hasselbalch equation is used which is \(pH = pKa + log \left( \frac {[Base]}{[Acid]} \right)\). Here, [Base] indicates the concentration of the base (Sodium formate) and [Acid] indicates the concentration of the acid (Formic acid). Substituting the given and found values in the equation we have, 3.50 = 3.74 + log \left( \frac {[Base]}{[Acid]} \right). Now we can solve for the ratio [Base]/[Acid].
04

Calculating concentrations

From the previous step, we can find the ratio [Base]/[Acid] = 0.58. Now to find the concentrations, one can keep the acid concentration at 0.100 M, and accordingly adjust the base concentration to achieve the required ratio. In this case, the base concentration will be 0.58 x 0.100 M = 0.058 M.
05

Finding the final volumes

Now, to prepare a buffer solution with the desired pH, you should mix the solutions in the right proportions. Since the sodium formate is 0.100 M and we need it to be 0.058 M, we take 0.58 L of sodium formate solution for each 1 L of formic acid solution. So preparing a buffer solution then depends on the total volume needed. If 1 L of buffer solution is needed, use 0.58 L of sodium formate and 1 L of formic acid. If less or more is needed, use the same ratios.

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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 powerful tool in chemistry, particularly when dealing with buffer solutions. This equation helps predict the pH of a buffer solution, which consists of a weak acid and its conjugate base. By using this formula, \[pH = pK_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)\]we can determine the pH based on the acid's dissociation constant, \(pK_a\), and the concentrations of the base and acid.

It's crucial because it directly relates the acid-base concentration ratio with the solution's pH. In practical applications, such as in laboratories or industries, adjusting the ratio of base to acid enables precise tuning of the pH to the desired level. This is particularly useful when the pH needs to remain stable, even if small amounts of an acid or base are added to the solution.

What's notable about using the Henderson-Hasselbalch equation is that it simplifies calculations. By knowing a base and its conjugate acid that fits the desired pH range, one can quickly adjust amounts to achieve exact pH control. Let's explore more about weak acids involved in these calculations next.
Weak Acid
A weak acid is a type of acid that only partially dissociates in water. This means that when dissolved, not all the acid molecules donate their hydrogen ions to water, resulting in an equilibrium between the undissociated and dissociated species. This characteristic affects how the acid behaves in a solution.

Weak acids remain mostly undissociated compared to strong acids, which completely ionize. Examples of weak acids include acetic acid (vinegar) and formic acid, which are common in everyday products or biological systems.

The concept of weak acids is vital in the context of buffers or equilibrium chemistry. Since they do not release all their hydrogen ions at once, they contribute to the stability of a buffer solution. In a buffer system made with a weak acid and its conjugate base, if an acidic or basic component is added, the solution can absorb these changes, helping maintain a nearly constant pH level.

Understanding weak acids, therefore, provides insight into how buffer systems work and how they help control environments within a narrow pH range, crucial for many biological and chemical processes.
pH Calculation
Calculating pH is a fundamental skill needed for understanding buffer solutions and acid-base chemistry. The pH, or 'potential of Hydrogen,' is a scale used to measure the acidity or basicity of an aqueous solution. It ranges from 0 to 14, with lower numbers indicating acidic solutions and higher numbers indicating basic solutions.

For any buffer solution, calculating its pH involves understanding how concentrations of the acid and its conjugate base work together. In buffer solutions, pH calculations often rely on the Henderson-Hasselbalch equation, which uses the relationship between the acid's \(pK_a\) and the molar concentrations of the acid and base.

Here's a simple guide to performing pH calculations in buffer solutions:
  • Identify the weak acid involved and look up its \(pK_a\).
  • Measure the initial concentrations of the acid and its conjugate base.
  • Apply the Henderson-Hasselbalch equation to find the pH.
This process is crucial in situations where maintaining a specific pH is necessary, such as in biological systems or chemical processes that are sensitive to changes in acidity.

By understanding how to calculate pH, you gain the ability to manipulate and stabilize solutions across various applications, ensuring optimal conditions for reactions or processes that depend on a specific pH.

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Most popular questions from this chapter

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?

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) is a somewhat stronger acid than water. Its ionization is represented by the equation \(\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HO}_{2}^{-}\) In \(1912,\) the following experiments were performed to obtain an approximate value of \(\mathrm{p} K_{\mathrm{a}}\) for this ionization at \(0^{\circ} \mathrm{C} .\) A sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was shaken together with a mixture of water and 1 -pentanol. The mixture settled into two layers. At equilibrium, the hydrogen peroxide had distributed itself between the two layers such that the water layer contained 6.78 times as much \(\mathrm{H}_{2} \mathrm{O}_{2}\) as the 1 -pentanol layer. In a second experiment, a sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was shaken together with 0.250 M NaOH(aq) and 1-pentanol. At equilibrium, the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was \(0.00357 \mathrm{M}\) in the 1-pentanol layer and 0.259 M in the aqueous layer. In a third experiment, a sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was brought to equilibrium with a mixture of 1 -pentanol and \(0.125 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq}) ;\) the concentrations of the hydrogen peroxide were \(0.00198 \mathrm{M}\) in the 1 -pentanol and \(0.123 \mathrm{M}\) in the aqueous layer. For water at \(0^{\circ} \mathrm{C}, \mathrm{p} K_{\mathrm{w}}=14.94\) Find an approximate value of \(\mathrm{pK}_{\mathrm{a}}\) for \(\mathrm{H}_{2} \mathrm{O}_{2}\) at \(0^{\circ} \mathrm{C}\) [Hint: The hydrogen peroxide concentration in the aqueous layers is the total concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) and \(\mathrm{HO}_{2}^{-}\). Assume that the 1 -pentanol solutions contain no ionic species.

Calculate the \(\mathrm{pH}\) at the points in the titration of \(25.00 \mathrm{mL}\) of 0.160 M HCl when (a) 10.00 mL and (b) 15.00 mL of 0.242 M KOH have been added.

Indicate whether you would expect the equivalence point of each of the following titrations to be below, above, or at \(\mathrm{pH}\) 7. Explain your reasoning. (a) \(\mathrm{NaHCO}_{3}(\mathrm{aq})\) is titrated with \(\mathrm{NaOH}(\mathrm{aq})\) (a) (b) \(\mathrm{HCl}(\mathrm{aq})\) is titrated with \(\mathrm{NH}_{3}(\mathrm{aq}) ;\) (c) \(\mathrm{KOH}(\mathrm{aq})\) is titrated with HI(aq).

What stoichiometric concentration of the indicated substance is required to obtain an aqueous solution with the pH value shown: (a) aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\), for \(\mathrm{pH}=8.95 ;(\mathrm{b}) \mathrm{NH}_{4} \mathrm{Cl}\) for \(\mathrm{pH}=5.12 ?\)

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