Related Problems with Solution
Problem 1 : Calculate the pH of a 0.02 M solution of NH4CN. Given Kb for NH4CN is $$(2.0 \times 10^{-5}) M.$$
Solution :
We can use the Kb expression for the ionization of NH4CN: $$[K_b = \frac{[NH_4^+][CN^-]}{[NH_4CN]}]$$
Since we have Kb and the initial concentration of NH4CN., 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 Kb 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.
