The image contains several chemistry problems in Khmer. I will solve question V (20 points). Given a 500 mL solution containing 0.06 mol of hydrofluoric acid (HF) and 0.06 mol of sodium fluoride (NaF), calculate: 1. The initial concentrations of HF and NaF in mol/L.

1. Problem Description

The image contains several chemistry problems in Khmer. I will solve question V (20 points).

Given a 500 mL solution containing 0.06 mol of hydrofluoric acid (HF) and 0.06 mol of sodium fluoride (NaF), calculate:

1. The initial concentrations of HF and NaF in mol/L.

2. The concentration of hydronium ions $[H_3O^+]$ and the pH of the solution.

3. Whether the solution is a buffer solution and why.

Given: $log\ 6.7 = 0.8$, $K_a(HF) = 6.7 \times 10^{-4}$.

2. Solution Steps

1. Calculate the initial concentrations of HF and NaF:

The volume of the solution is 500 mL = 0.5 L.

The number of moles of HF is 0.06 mol.

The number of moles of NaF is 0.06 mol.

$[HF] = \frac{0.06\ mol}{0.5\ L} = 0.12\ M$

$[NaF] = \frac{0.06\ mol}{0.5\ L} = 0.12\ M$

2. Calculate the concentration of hydronium ions and the pH of the solution.

The equilibrium reaction is:

$HF(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + F^-(aq)$

$K_a = \frac{[H_3O^+][F^-]}{[HF]}$

Since NaF is a salt that completely dissociates in water:

$NaF(aq) \rightarrow Na^+(aq) + F^-(aq)$

Therefore, $[F^-] \approx [NaF]$

We can use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution:

$pH = pK_a + log \frac{[F^-]}{[HF]}$

$pK_a = -log\ K_a = -log(6.7 \times 10^{-4}) = -log\ 6.7 - log\ 10^{-4} = -0.8 + 4 = 3.2$

Since $[HF] = [F^-]$, the log term becomes:

$log \frac{[F^-]}{[HF]} = log\ 1 = 0$

Therefore, $pH = pK_a = 3.2$

Now we can calculate the hydronium ion concentration:

$[H_3O^+] = 10^{-pH} = 10^{-3.2} = 10^{-3.2} = 10^{0.8} \times 10^{-4} = 6.3 \times 10^{-4}\ M$

(Since log 6.7= 0.8, the antilog of 0.8 = 6.

3. The value is slightly different than 6.7 due to rounding error.)

3. Determine if the solution is a buffer solution.

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added. It typically consists of a weak acid and its conjugate base, or a weak base and its conjugate acid.

In this case, we have a solution containing a weak acid (HF) and its conjugate base (F-), provided by NaF. Therefore, this solution is a buffer solution.

3. Final Answer

1. $[HF] = 0.12\ M$, $[NaF] = 0.12\ M$

2. $[H_3O^+] = 6.3 \times 10^{-4}\ M$, $pH = 3.2$

3. Yes, the solution is a buffer solution because it contains a weak acid (HF) and its conjugate base (F-).