Exemplar Problems
Problem 9 : Determine the pH of a 0.02 M solution of NH4CN. Given Kb for NH4CN is $$(2.0 \times 10^{-5}) M.$$
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 Kb, which is the equilibrium constant for the reaction of NH4CN with water to produce NH3 and CN-:
$$[NH_4CN + H_2O \rightleftharpoons NH_3 + CN^- + H_3O^+]$$
The expression for Kb is as follows:
$$[K_b = \frac{[NH_3][CN^-]}{[NH_4CN]}]$$
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: $$[NH_4CN] - (x) M$$ $$[NH_3] + (x) M$$ $$[CN^-] + (x) M$$
Using the Kb 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):
$$[pH = 14 - pOH = 14 - (-\log(2.0 \times 10^{-4}))]$$
Calculate (pH):
$$[pH \approx 9.70]$$
So, the pH of the 0.02 M solution of NH4CN is approximately 9.70.
BETA
