Question

question 17 (4 points)
the ka of hydrofluoric acid (hf) at 25°c is 6.8 x 10^-4. what is the ph of a 0.45m aqueous solution of hf?

a) 4.05
b) 1.76
c) 3.64
d) 1.41

Explanation:

Step1: Write the dissociation equation

Hydrofluoric acid (HF) dissociates in water as: $HF
ightleftharpoons H^+ + F^-$
Let the concentration of $H^+$ ions at equilibrium be $x$. Then the concentration of $F^-$ ions will also be $x$ (from the 1:1 dissociation), and the concentration of undissociated HF will be $0.45 - x$.

Step2: Write the Ka expression

The acid dissociation constant $K_a$ is given by:
$$K_a = \frac{[H^+][F^-]}{[HF]}$$
Substituting the equilibrium concentrations:
$$6.8 \times 10^{-4} = \frac{x \cdot x}{0.45 - x}$$

Step3: Approximate (check if $x$ is small)

Since $K_a$ is relatively small ($6.8 \times 10^{-4}$) and the initial concentration of HF ($0.45\ M$) is relatively large, we can approximate $0.45 - x \approx 0.45$ (this is valid if $x \ll 0.45$, which we can check later). So the equation becomes:
$$6.8 \times 10^{-4} = \frac{x^2}{0.45}$$

Step4: Solve for $x$

Multiply both sides by $0.45$:
$$x^2 = 6.8 \times 10^{-4} \times 0.45$$
$$x^2 = 3.06 \times 10^{-4}$$
Take the square root of both sides:
$$x = \sqrt{3.06 \times 10^{-4}} \approx 0.0175\ M$$
Now check the approximation: $\frac{0.0175}{0.45} \times 100 \approx 3.89\%$, which is less than $5\%$, so the approximation is valid.

Step5: Calculate pH

The pH is given by $pH = -\log[H^+]$. Since $[H^+] = x \approx 0.0175\ M$:
$$pH = -\log(0.0175) \approx 1.76$$

Answer:

b) 1.76