(a) Explain what is meant by a weak acid.

Methanoic acid (HCO2H) has a numerical value of Ka(T) at 298K of 1.60 × 10-4
(b) Write an expression for Ka(T) of methanoic acid, and state its units.
(c) What is the pH value of a 0.10 mol dm-3 aqueous solution of methanoic acid? Give a clear account of your calculation and state any assumptions you have made

Weak acids don't ionize completely.

HCOOH ==> H^+ + HCOO^- for which
Ka = (H^+)(HCOO^-)/(HCOOH)
There are no unit for Ka. If you use molarity for concn, as most do, there appears to be "formal" units but in higher chemistry courses you will learn there are no units since the numbers that go in are activities and they have no units.

What is the pH value of a 0.10 mol dm-3 aqueous solution of methanoic acid? Give a clear account of your calculation and state any assumptions you have made

(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.