Problem 30 - Chemistry: An Atoms-Focused Approach

Question

What masses of ethylamine and ethylammonium chloride do you need to prepare \(1.00 \mathrm{L}\) of \(\mathrm{pH}=10.50\) buffer if the total concentration of the two components is \(0.250 M ?\)

Step-by-Step Solution

Answer

Answer: To prepare 1.00 L of pH 10.50 buffer solution, you will need approximately 4.06 g of ethylamine and 8.56 g of ethylammonium chloride.

Step 1: Write the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation for a buffer system is given as: \(pH = pK_a + \log \frac{[\text{Base}]}{[\text{Acid}]}\)

Step 2: Find the equilibrium dissociation constant (Ka) of ethylamine

Ethylamine is a weak base and its conjugate acid is ethylammonium ion. We need to find the \(K_b\) value to use in the Henderson-Hasselbalch equation. The \(K_b\) value for ethylamine is \(5.6 \times 10^{-4}\). To find the \(K_a\) value for the conjugate acid, we use the relationship between \(K_w\), \(K_a\), and \(K_b\): \(K_w = K_a \times K_b = 1.0 \times 10^{-14}\) Solving for \(K_a\) we get: \(K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{5.6 \times 10^{-4}} = 1.79 \times 10^{-11}\)

Step 3: Calculate the pKa value

Now that we know the \(K_a\) value, we can find the \(pK_a\) value as follows: \(pK_a = -\log K_a = -\log(1.79 \times 10^{-11}) = 10.75\)

Step 4: Find the ratio of Base to Acid

Using the pH and pKa value in the Henderson-Hasselbalch equation, we can find the ratio of Base to Acid: \(10.50 = 10.75 + \log \frac{[\text{Base}]}{[\text{Acid}]}\) Solving for the ratio, we get: \(\frac{[\text{Base}]}{[\text{Acid}]} = 10^{10.50 - 10.75} = 10^{-0.25} \approx 0.56\)

Step 5: Determine the concentrations of the components

The total concentration of the two components is given as 0.250 M. Let \(x\) be the concentration of the Acid (ethylammonium chloride) and \(y\) be the concentration of the Base (ethylamine). We have two equations: 1. \(x + y = 0.250 M\) 2. \(\frac{y}{x} = 0.56\) By solving these equations, we get: \(x = \frac{0.250}{1 + 0.56} \approx 0.160\,M\) \(y = 0.250 - 0.160 = 0.090\,M\)

Step 6: Calculate the masses of the components

We will now find the masses of ethylamine and ethylammonium chloride needed to prepare 1.00 L of the buffer solution, using the calculated concentrations and their molar masses (M\(_1\) and M\(_2\)): Mass of ethylamine \(= y \times \text{Volume} \times \text{M}_1 = (0.090\,\text{M}) \times (1.00\,\text{L}) \times (45.08\,\text{g/mol}) \approx 4.06\,\text{g}\) Mass of ethylammonium chloride \(= x \times \text{Volume} \times \text{M}_2 = (0.160\,\text{M}) \times (1.00\,\text{L}) \times (53.50\,\text{g/mol}) \approx 8.56\,\text{g}\) To prepare 1.00 L of pH 10.50 buffer, you will need approximately 4.06 g of ethylamine and 8.56 g of ethylammonium chloride.