(a) A weak acid is an acid that does not completely dissociate in water. When a weak acid is dissolved in water, only a small fraction of the acid molecules ionize to form H+ ions. The rest of the acid molecules remain in their undissociated form. As a result, a weak acid has a lower concentration of H+ ions compared to a strong acid.
(b) The expression for Ka(T) of methanoic acid can be written as:
Ka(T) = [H+][HCO2-] / [HCO2H]
The units of Ka(T) depend on the concentration units used. In this case, the concentration is given in mol dm-3, so the units of Ka(T) would be (mol dm-3) / (mol dm-3) = dm3 mol-1.
(c) To calculate the pH value of a 0.10 mol dm-3 aqueous solution of methanoic acid, we can make use of the Ka(T) value given and the assumption that the concentration of H+ ions is equal to the concentration of HCO2- ions (due to the 1:1 stoichiometry).
First, we need to set up an equation to determine the concentration of H+ ions. Let x be the concentration of H+ ions, which is also the concentration of HCO2- ions. The concentration of undissociated methanoic acid (HCO2H) can be calculated as (0.10 - x) mol dm-3.
Using the Ka(T) expression, we can write:
1.60 × 10-4 = x^2 / (0.10 - x)
Since x is small compared to 0.10, we can assume that (0.10 - x) ≈ 0.10.
1.60 × 10-4 = x^2 / 0.10
Rearranging the equation, we get:
x^2 = 0.10 * 1.60 × 10-4
x^2 = 1.60 × 10-5
Taking the square root of both sides, we find:
x ≈ 1.26 × 10-3 mol dm-3
Now that we have the concentration of H+ ions, we can calculate the pH using the formula:
pH = -log[H+]
pH ≈ -log(1.26 × 10-3)
pH ≈ 2.90
Therefore, the pH value of a 0.10 mol dm-3 aqueous solution of methanoic acid is approximately 2.90.