What is the pH a 0.25 mol/L solution of methanoic acid
The pH of a 0.25 mol/L solution of methanoic acid can be determined using the dissociation constant (Ka) of methanoic acid, which is 1.8 x 10^-4 at 25°C. Methanoic acid (also known as formic acid) is a weak acid, meaning it only partially dissociates in water.
To calculate the pH, we use the formula: pH = -log10[H+]
First, we need to determine the concentration of H+ ions in the solution. Since methanoic acid partially dissociates according to the equation:
CHOOH (methanoic acid) ⇌ H+ + CHO2- (methanoate ion)
The concentration of H+ ions produced can be considered equal to the concentration of methanoic acid that dissociates.
Therefore, for every mole of methanoic acid that dissociates, one mole of H+ ions is produced.
In a 0.25 mol/L solution of methanoic acid, 0.25 mol of methanoic acid will dissociate:
[H+] = 0.25 mol/L
Now, we can calculate the pH:
pH = -log10(0.25)
= 0.60
Therefore, the pH of a 0.25 mol/L solution of methanoic acid is approximately 0.60.