Explain the difference, with sketches, between the titration curves of monoprotic acid and a triprotic acid.

A monoprotic acid has only one equivalence point in a titration curve while a triprotic acid has three equivalence points.

Explanation:

Monoprotic acids, HA, are acids that are able to donate one proton per molecule during the process of ionization or dissociation as shown below:

HA ----> H⁺ + A⁻

A triprotic acid is a polyproprotic acid, that is acids that can donate more than one proton per molecule during ionization. a triprotic acid can donate three protons per molecule during the process of ionization. The dissociation of a triprotic acid is shown below;

H₃A -----> H₂A⁻ + H⁺

H₂A⁻ ----> HA²⁻ + H⁺

HA²⁻ -----> A³⁻ + H⁺

The value Ka, is known as the acid dissociation constant. It is found by looking at the equilibrium constant for the dissociation of the acid. The higher the Ka, the more the acid dissociates

A titration curve is the plot of the pH of the analyte solution versus the volume of the titrant added as the titration progresses.

In a titration curve in which base is added to acid, the pH rises slowly until the equivalent point at which point the pH remains constant before it begins to rise again on the addition of more base.

In a titration curve of a monoprotic acid, there is only one region in which the equivalence point is obtained corresponding to the point of acid of dissociation of all its protons.

In the titration curve of a triproprotic acid, three equivalent points are obtained corresponding to the point of dissociation of each of three protons of the triprotic acid.

The titration curves are shown below

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g