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How many moles of hydrofluoric acid, HF, must be added to \(500.0 \mathrm{~mL}\) of \(0.25 M\) sodium fluoride to give a buffer of \(\mathrm{pH} 3.50 ?\) Ignore the volume change due to the addition of hydrofluoric acid.

Short Answer

Expert verified
Add 0.0585 moles of HF to the solution.

Step by step solution

01

Identify Components of the Buffer Solution

In this exercise, we have a buffer solution consisting of hydrofluoric acid (HF) and its conjugate base, sodium fluoride (NaF). The purpose of a buffer is to maintain a relatively constant pH.
02

Use the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is used for buffer solutions as follows: \[\text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]\right)\] where \([\text{A}^-]\) is the concentration of the conjugate base (NaF, which provides \(\text{F}^-\)), and \([\text{HA}]\) is the concentration of the acid (HF).
03

Determine the pKa of Hydrofluoric Acid

The dissociation constant \(K_a\) of HF is approximately \(6.8 \times 10^{-4}\). Therefore, the \(\text{pK}_a\) is:\[\text{pK}_a = -\log(K_a) = -\log(6.8 \times 10^{-4}) \approx 3.17\]
04

Set Up the Equation with Given pH

Place the given pH and calculated \(\text{pK}_a\) into the Henderson-Hasselbalch equation:\[3.50 = 3.17 + \log\left(\frac{0.25}{[\text{HF}]\right)\]Rearrange to solve for \([\text{HF}]\):\[0.33 = \log\left(\frac{0.25}{[\text{HF}]\right)\]
05

Solve for the Concentration of HF

Exponentiate both sides to solve for \([\text{HF}]\):\[10^{0.33} = \frac{0.25}{[\text{HF}]}\]Calculate:\[2.14 \approx \frac{0.25}{[\text{HF}]}\]Therefore,\[[\text{HF}] \approx \frac{0.25}{2.14} \approx 0.117 \text{ M}\]
06

Calculate Moles of HF Added

The solution volume is 500.0 mL or 0.500 L. Use the relationship \(\text{moles} = \text{molarity} \times \text{volume}\) to convert the molarity to moles:\[\text{moles of HF} = 0.117 \text{ M} \times 0.500 \text{ L} = 0.0585 \text{ moles}\]

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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 formula used to relate the pH of a buffer solution to the concentration of the acid and its conjugate base. It is expressed as:\[\text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\]
  • \(\text{pH}\) represents the acidity or alkalinity of the solution.
  • \(\text{pK}_a\) is the negative logarithm of the acid dissociation constant \(K_a\).
  • \([\text{A}^-]\) is the concentration of the conjugate base.
  • \([\text{HA}]\) is the concentration of the acid.
Using this equation, we can calculate either pH, \([\text{A}^-]\), or \([\text{HA}]\) if the other values are known. It is extremely helpful for working with buffer solutions because buffers aim to resist changes in pH when acids or bases are added. Remember, precision in your calculations is key to obtaining accurate results.
pH Calculation
Calculating the pH of a solution is a fundamental skill in chemistry. The pH is determined by the concentration of hydrogen ions in the solution. For buffer solutions, as in our exercise, the pH is calculated using the Henderson-Hasselbalch equation.Let's apply it: For the given buffer solution containing hydrofluoric acid (HF) and sodium fluoride (NaF), the aim is to find out how much HF is needed to achieve a specific pH. After determining the \(\text{pK}_a\) of HF and knowing the concentration of the conjugate base (\([\text{F}^-]\)), you can plug the values into the equation and solve for the concentration of HF (\([\text{HA}]\)).The steps usually follow:
  • Insert the known \(\text{pH}\) and \(\text{pK}_a\) into the equation.
  • Plug in the known concentration of \([\text{A}^-]\).
  • Solve the equation for \([\text{HA}]\) to find out the needed concentration.
This calculated concentration of the acid informs you of how much needs to be added to the solution to reach the desired pH.
pKa Value Calculation
The \(\text{pK}_a\) value is essential for determining the strength of an acid within a buffer solution. It represents the negative logarithm of the acid's dissociation constant \(K_a\), providing insights into the degree of ionization of the acid in solution.To calculate \(\text{pK}_a\):
  • Find the \(K_a\) value of the acid, which is often found in reference tables.
  • Use the formula: \(\text{pK}_a = -\log(K_a)\).
For hydrofluoric acid (HF), the \(K_a\) is given as \(6.8 \times 10^{-4}\).
Applying the formula, \(\text{pK}_a = -\log(6.8 \times 10^{-4}) \approx 3.17\).
This \(\text{pK}_a\) value plays a critical role in the Henderson-Hasselbalch equation, helping to determine the pH of the buffer solution formed by the acid and its conjugate base.
Molarity and Moles Conversion
Molarity is a concentration measurement expressed as moles of solute per liter of solution (mol/L). It is often denoted by the letter \(M\). Understanding how to convert between molarity and moles is crucial for solving problems like the one in the exercise.Here’s a simple guide on how to do this:
  • Find the molarity (concentration) from step 5.
  • Convert volume from milliliters to liters if needed (e.g., 500.0 mL = 0.500 L).
  • Use the conversion formula: \(\text{moles} = \text{molarity} \times \text{volume}\).
In the given exercise, with \(0.117\, \text{M}\) HF and a volume of \(0.500\, \text{L}\), the moles of HF needed can be calculated as:
\[0.117 \times 0.500 = 0.0585\, \text{moles}\].
Always ensure that units are consistent when calculating, and you will be able to interpret these figures to manipulate and understand the chemistry of buffer solutions effectively.

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Most popular questions from this chapter

A solution of acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\), on a laboratory shelf was of undetermined concentration. If the pH of the solution was found to be \(2.68\), what was the concentration of the acetic acid?

a. For each of the following salts, write the reaction that occurs when it dissociates in water: \(\mathrm{NaCl}(s), \mathrm{NaCN}(s)\), \(\mathrm{KClO}_{2}(s), \mathrm{NH}_{4} \mathrm{NO}_{3}(s), \operatorname{KBr}(a q)\), and \(\mathrm{NaF}(s)\) b. Consider each of the reactions that you wrote above, and identify the aqueous ions that could be proton donors (acids) or proton acceptors (bases). Briefly explain how you decided which ions to choose. c. For each of the acids and bases that you identified in part b, write the chemical reaction it can undergo in aqueous solution (its reaction with water). d. Are there any reactions that you have written above that you anticipate will occur to such an extent that the \(\mathrm{pH}\) of the solution will be affected? As part of your answer, be sure to explain how you decided. e. Assume that in each case above, \(0.01 \mathrm{~mol}\) of the salt was dissolved in enough water at \(25^{\circ} \mathrm{C}\) to make \(1.0 \mathrm{~L}\) of solution. In each case, what additional information would you need in order to calculate the \(\mathrm{pH}\) ? If there are cases where no additional information is required, be sure to state that as well. f. Say you take \(0.01\) mol of \(\mathrm{NH}_{4} \mathrm{CN}\) and dissolve it in enough water at \(25^{\circ} \mathrm{C}\) to make \(1.0 \mathrm{~L}\) of solution. Using chemical reactions and words, explain how you would go about determining what effect this salt will have on the \(\mathrm{pH}\) of the solution. Be sure to list any additional information you would need to arrive at an answer.

Find the \(\mathrm{pH}\) of the solution obtained when \(25 \mathrm{~mL}\) of \(0.065 M\) benzylamine, \(\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{NH}_{2}\), is titrated to the equivalence point with \(0.050 M\) hydrochloric acid. \(K_{b}\) for benzylamine is \(4.7 \times 10^{-10}\)

Calculate the \(\mathrm{pH}\) ot a solution made by mixing \(7.52 \mathrm{~mL}\) of \(4.9 \times 10^{-2} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) with \(22.5 \mathrm{~mL}\) of \(0.11 \mathrm{M} \mathrm{HCl}\)

If \(20.0 \mathrm{~mL}\) of a \(0.10 \mathrm{M} \mathrm{NaOH}\) solution is added to a \(30.0\) mL sample of a \(0.10 M\) weak acid, HA, what is the \(\mathrm{pH}\) of the resulting solution? \(\left(K_{a}=1.8 \times 10^{-5}\right.\) for HA \()\) a. \(2.87\) b. \(2.74\) c. \(4.74\) d. \(5.05\) e. \(8.73\)

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