Exemplar Problems

Problem 6 : Calculate the pH of a solution prepared by mixing 100 mL of 0.1 M $HCl$ and 100 mL of 0.1 M NH₃OH₊

Solution :

Given that K_w (the ion product of water) is $$(1.0 \times 10^{-14})$$ at 25°C, and we have a solution of (HCl) with a H⁺ concentration of $$(1.0 \times 10^{-3}) M,$$ we can calculate the OH^- concentration using K_w:

$$[K_w = [H^+][OH^-]]$$

Substitute the given [H⁺] concentration:

$$(1.0 \times 10^{-14} = (1.0 \times 10^{-3})([OH^-]))$$

Solve for OH^-:

$$[ [OH^-] = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-3}}]$$

$$[ [OH^-] = 1.0 \times 10^{-11} \text{ M}]$$

Now, we can calculate the (pOH):

$$[pOH = -\log[OH^-] = -\log(1.0 \times 10^{-11})]$$

$$[pOH \approx 11]$$

To find the pH, we can use the fact that $$(pH + pOH = 14):$$

$$[pH = 14 - pOH = 14 - 11 = 3]$$

So, the pH of the solution is 3.