### Related Problems with Solution

##### Problem 2 : Calculate the pH of a 0.1 M solution of (NaF). Given K_{a} for (HF) is $$(6.8 \times 10^{-4}).$$

##### Solution :

The ionization of (NaF) in water is as follows: $$[NaF \rightleftharpoons Na^+ + F^-]$$

We can use the K_{a} expression for the ionization of (HF):
$$[K_a = \frac{[H^+][F^-]}{[HF]}]$$

Since (NaF) is a salt, it dissociates completely into its ions. So, the concentration of F^{-} ions from (NaF) is 0.1 M.

Now, we can set up an ICE table for the ionization of (HF):

```
HF => H+ + F-
----------------------------
Initial 0 M 0 M 0 M
Change -x x x
Equilibrium -x x x
```

From the K_{a} expression, we have:
$$[6.8 \times 10^{-4} = \frac{x \cdot x}{0.1 - x}]$$

Since (x) is small compared to 0.1, we can approximate (0.1 - x) as 0.1: $$[6.8 \times 10^{-4} = \frac{x^2}{0.1}]$$

Now, solve for (x): $$[x^2 = 6.8 \times 10^{-4} \cdot 0.1]$$ $$[x^2 = 6.8 \times 10^{-5}]$$ $$[x = \sqrt{6.8 \times 10^{-5}}]$$ $$[x \approx 8.26 \times 10^{-3} , \text{M}]$$

Now, calculate the pH using the H^{+} ion concentration:
$$[pH = -\log(8.26 \times 10^{-3})]$$

Calculate (pH): $$[pH \approx 2.08]$$

So, the pH of the solution is approximately 2.08.