Define (i) a base, (ii) a conjugate acid-base pair, according to Brønsted-Lowry theory - Leaving Cert Chemistry - Question 9 - 2017

A conjugate acid-base pair consists of two species that differ by the presence of a single proton (H⁺). For example, NH₃ (ammonia) and NH₄⁺ (ammonium) are a conjugate acid-base pair.

The conjugate acid of H₂O (water) is the hydronium ion (H₃O⁺), formed when H₂O accepts a proton.

pH is a measure of the acidity or alkalinity of a solution. It is defined mathematically as:

$pH = - ext{log}_{10}([H^+])$

where [H⁺] represents the concentration of hydrogen ions in moles per liter.

Hydrochloric acid (HCl) is a strong monoprotic acid that fully dissociates in water, producing 0.10 M H⁺ ions. Sulfuric acid (H₂SO₄) is a diprotic acid, and its first dissociation produces two moles of H⁺ for every mole of acid, resulting in a concentration of 0.20 M H⁺ for 0.10 M H₂SO₄, leading to a lower pH.

Methanoic acid (HCOOH) is a weak acid that only partially dissociates in solution. As a result, at the same concentration, it will produce a lower concentration of H⁺ ions compared to hydrochloric acid, yielding a higher pH value.

Given the pH of 2.37 for the methanoic acid solution, we can calculate [H⁺]:

$ext{[H⁺]} = 10^{-2.37} = 4.27 imes 10^{-3} ext{ M}$

Using the expression for Ka: K_a = rac{[ ext{H⁺}][ ext{HCOO}^-]}{[ ext{HCOOH}]} Assuming initial concentration of HCOOH is 0.10 M, we can approximate: K_a = rac{(4.27 imes 10^{-3})^2}{0.10 - 4.27 imes 10^{-3}} ext{ and solve for } K_a

For the 0.05 M methanoic acid solution, the equilibrium expression is: K_a = rac{[ ext{H⁺}]^2}{0.05 - [ ext{H⁺}]} By substituting the calculated Ka value and solving for [H⁺], we can then find the corresponding pH using: $pH = - ext{log}_{10}([ ext{H⁺}])$

In the pH curve diagram, the x-axis should represent the volume of NaOH added, while the y-axis represents the pH. The curve will show a gradual increase in pH with a steep rise around the equivalence point where neutralization occurs, typically between 7 and 8 pH.

A suitable indicator for this titration would be phenolphthalein, because it changes color in the pH range that corresponds to the equivalence point of the titration, which is typically around 8 to 10.

Phenolphthalein is chosen due to its color change occurring at around pH 8 to 10, which coincides with the sharp rise in pH that would be observed on the pH curve. This allows for accurate determination of the endpoint of the titration.