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 NH4OH
Solution :
Given that Kw (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 Kw:
$$[K_w = [H^+][OH^-]]$$
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.
BETA
