Problem 58 - Chemistry: The Central Science

Question

How many moles of \(\mathrm{HF}\left(K_{a}=6.8 \times 10^{-4}\right)\) must be present in \(0.200 \mathrm{~L}\) to form a solution \(w\) ith a \(\mathrm{pH}\) of \(3.25\) ?

Step-by-Step Solution

Answer

To form a 0.200 L solution with a pH of 3.25 using HF with a given \(K_{a} = 6.8 \times 10^{-4}\), there must be \(9.36 \times 10^{-5}\) moles of HF present in the solution.

Step 1: Find the concentration of H+ ions

We can find the concentration of hydrogen ions \([\mathrm{H^+}]\) in the solution using the provided pH value. Recall the relationship between pH and \([\mathrm{H^+}]\): \( \mathrm{pH} = - \log [\mathrm{H^+}] \) Given the pH value of 3.25, we can calculate the hydrogen ion concentration. \[ [\mathrm{H^+}] = 10^{-\mathrm{pH}} \] \[ [\mathrm{H^+}] = 10^{-3.25} \] Now, we need to calculate the concentration of H+ ions: \[ [\mathrm{H^+}] = 5.62 \times 10^{-4} \] So, the concentration of hydrogen ions in the solution is \(5.62 \times 10^{-4} M\).

Step 2: Set up equilibrium expression for Ka

We know the acid dissociation constant, Ka, for HF. We can set up an equilibrium expression considering the dissociation of HF: \( \mathrm{HF} \rightleftharpoons \mathrm{H^+} + \mathrm{F^-} \) The Ka expression for this reaction is: \[K_{a} = \frac{[\mathrm{H^+}][\mathrm{F^-}]}{[\mathrm{HF}]} \] Since the concentrations of dissociated H+ and F- ions will be equal and we are trying to find the amount of undissociated HF, we can denote the concentration of HF as x. So, the expression becomes: \[6.8 \times 10^{-4} = \frac{ (5.62 \times 10^{-4})^2 }{ x } \]

Step 3: Calculate the concentration of HF

Now we can solve for x, which is the concentration of HF in the solution. \[x = \frac{ (5.62 \times 10^{-4})^2}{6.8 \times 10^{-4}} \] \[x = 4.68 \times 10^{-4} \] So, the concentration of HF in the solution is \(4.68 \times 10^{-4} M\).

Step 4: Determine the number of moles of HF

Now that we know the concentration of HF in the 0.200 L solution, we can calculate the number of moles of HF required. Recall the relationship between moles (n), concentration (c), and volume (V): \[n = c \times V \] We can plug in the values we have found: \[n = (4.68 \times 10^{-4} \mathrm{M}) \times (0.200 \mathrm{L}) \] \[n = 9.36 \times 10^{-5} \] So, there must be \(9.36 \times 10^{-5}\) moles of HF present in the 0.200 L solution to achieve a pH of 3.25.

Key Concepts

pH calculation

Understanding pH calculation is a fundamental part of grasping acid-base equilibrium. pH is a scale that measures how acidic or basic a solution is. It ranges from 0 to 14, with lower numbers indicating stronger acids and higher numbers indicating stronger bases.

To find the pH, you need to know the concentration of hydrogen ions \(\left([\text{H}^+]\right)\) in the solution. The formula is \( ext{pH} = -\log [\text{H}^+]\). This logarithmic function means that each pH unit change corresponds to a tenfold change in hydrogen ion concentration.

In our example, we are given a pH of 3.25. By rearranging the formula, we can find the concentration of \([\text{H}^+]\): \[ [\text{H}^+] = 10^{-\text{pH}} \]. Substituting the pH value, we calculate \[ [\text{H}^+] = 10^{-3.25} = 5.62 \times 10^{-4} \].

• This means the solution has a hydrogen ion concentration of \([\text{H}^+] = 5.62 \times 10^{-4}\) M.
• Knowing this concentration is essential because it helps us understand the degree of acidity of the solution.

acid dissociation constant

In the world of chemistry, the acid dissociation constant, usually represented as \(K_a\), is a value that quantifies the strength of an acid in a solution. It helps us understand how well an acid dissociates into ions, influencing the acidity.

For an acid \( \text{HA}\), dissociating in water can be represented as: \[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \].The \(K_a\) expression for this dissociation is \[ K_{a} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \].

In this problem, we applied this concept to hydrofluoric acid \( \text{HF} \). Its dissociation is given by: \[ \text{HF} \rightleftharpoons \text{H}^+ + \text{F}^- \], and the expression becomes: \[ K_{a} = \frac{[\text{H}^+][\text{F}^-]}{[\text{HF}]} \]. Given \(K_a = 6.8 \times 10^{-4}\), we set up an equation with known \([\text{H}^+]\) from the pH calculation.

• Because \([\text{H}^+]\) and \([\text{F}^-]\) are equal at equilibrium, we use \(5.62 \times 10^{-4}\) to solve for \(x\), the concentration of undissociated HF.
• This calculation yields the concentration \(x = 4.68 \times 10^{-4}\) M.

mole calculation

Calculating moles from a given concentration and volume is a crucial skill in chemistry, particularly when determining how much of a substance is needed for a certain chemical condition.

The formula \(n = c \times V\) links moles (), concentration (c,), and volume (V). In this case, we aim to figure out how many moles of HF are necessary to create a specified pH level in a particular volume.

We've got the concentration of HF from the previous section: \(4.68 \times 10^{-4} \, \text{M}\). Given that the volume of the solution is 0.200 L, substitute these into our formula to find, \[ n = (4.68 \times 10^{-4} \, \text{M}) \times (0.200 \, \text{L}) \].

• This gives us \( n = 9.36 \times 10^{-5} \) moles.
• Therefore, to achieve the pH of 3.25 in 0.200 L of solution, we need \(9.36 \times 10^{-5}\) moles of HF.

Understanding this relationship allows you to effectively prepare solutions of desired properties by calculating how much solute is needed.