### Related Problems with Solution

##### Problem 1 : Calculate the pH of a 0.02 M solution of NH_{4}CN. Given K_{b} for NH_{4}CN is $$(2.0 \times 10^{-5}) M.$$

##### Solution :

We can use the K_{b} expression for the ionization of NH_{4}CN:
$$[K_b = \frac{[NH_4^+][CN^-]}{[NH_4CN]}]$$

Since we have K_{b} and the initial concentration of NH_{4}CN., we can set up an ICE (initial, change, equilibrium) table to calculate the concentration of (OH^-) ions and then find (pOH):

```
NH4CN => NH4+ + CN-
---------------------------------
Initial 0.02 M 0 M 0 M
Change -x x x
Equilibrium 0.02 - x x x
```

From the K_{b} expression, we have:
$$[2.0 \times 10^{-5} = \frac{x \cdot x}{0.02 - x}]$$

Since (x) is small compared to 0.02, we can approximate (0.02 - x) as 0.02: $$[2.0 \times 10^{-5} = \frac{x^2}{0.02}]$$

Now, solve for (x): $$[x^2 = 2.0 \times 10^{-5} \cdot 0.02]$$ $$[x^2 = 4.0 \times 10^{-7}]$$ $$[x = \sqrt{4.0 \times 10^{-7}}]$$ $$[x = 2.0 \times 10^{-4} , \text{M}]$$

Now that we have the OH^{-} ion concentration, we can calculate (pOH):
$$[pOH = -\log(2.0 \times 10^{-4})]$$

Calculate (pOH): $$[pOH \approx 3.70]$$

Finally, we can find the pH using the relation: (pH + pOH = 14): $$[pH = 14 - pOH]$$ $$[pH \approx 14 - 3.70]$$ $$[pH \approx 10.30]$$

So, the pH of the solution is approximately 10.30.