Exemplar Problems

Problem 9 : Determine the pH of a 0.02 M solution of NH₄CN. Given K_b for NH₃ is $$(1.8 \times 10^{-5}) M.

Solution :

To determine the pH of a solution of NH₄CN, we first need to calculate the concentration of OH^- ions in the solution using the given value of K_b, which is the equilibrium constant for the reaction of CN^- with water to produce NH₃ and NH₄⁺:

$$[NH_4CN + H_2O \rightleftharpoons NH_4^+ + CN^- + OH^-]$$

The expression for K_b is as follows:

$$[K_b = \frac{[NH_4^+][CN^-]}{[NH_4CN]}]$$

Given that $$(K_b = 2.0 \times 10^{-5}) M$$ and the initial concentration of NH₄CN is 0.02 M, we can set up an ICE (initial, change, equilibrium) table:

Initial: $$[NH_4CN] = 0.02 M$$ $$[NH_3] = 0 M$$ $$[CN^-] = 0 M$$

Change: $$[NH_4CN] decreases by (-x)$$ $$[NH_3] increases by (x)$$ $$[CN^-] increases by (x)$$

Equilibrium: $$[NH_4CN] - (x) M$$ $$[NH_3] + x\ \text{M}$$ $$[CN^-] + (x) M$$

Using the K_b expression:

$$[2.0 \times 10^{-5} = \frac{(x)(x)}{0.02 - x}]$$

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):

$$[x^2 = 2.0 \times 10^{-5} \times 0.02]$$ $$[x^2 = 4.0 \times 10^{-7}]$$ $$[x = \sqrt{4.0 \times 10^{-7}}]$$ $$[x \approx 2.0 \times 10^{-4}]$$

Now that we have the concentration of OH^- ions, we can calculate the (pOH):

$$[pOH = -\log[OH^-] = -\log(2.0 \times 10^{-4})]$$

Now, calculate the pH using the fact that (pH + pOH = 14):

$$[pH = 14 - pOH = 14 - (-\log(2.0 \times 10^{-10}))]$$

Calculate (pH):

$$[pH \approx 9.70]$$

So, the pH of the 0.02 M solution of NH₄CN is approximately 8.70.