Astudent was given a 50.0 ml of 0.10 m solution of an unknown diprotic acid, h2a, which was titrated with 0.10 m naoh. after a total of 25.0 ml of naoh was added, the resulting solution has a ph of 6.70.
after a total of 50.0 ml of naoh was added, the ph of the solution was 8.00.

determine the values of ka1 and ka2 for the diprotic acid.

Ka1 = 2.00x10⁻⁷, Ka2 = 5.00x10⁻¹⁰

Explanation:

A diprotic acid is a substance that can release 2 H⁺ when in aqueous solution. Because it is a weak acid, the ionization will be reversible. So, the acid has two equilibrium reactions, each one with its equilibrium constant:

H₂A ⇄ H⁺ + HA⁻   Ka1 = ([HA⁻]*[H⁺]/[H₂A])

HA⁻ ⇄ H⁺ + A⁻     Ka2 = ([A⁻]*[H⁺]/[HA⁻])

First, the number of moles of H₂A was:

n = 0.10 mol/L *0.05L = 0.005 mol

And then was added NaOH:

n = 0.1 mol/L * 0.025 L = 0.0025 mol

So, all the NaOH will reacts, then, the number of moles will be:

H₂A = 0.005 - 0.0025 = 0.0025 mol

HA⁻ = 0.0025 mol (the stoichiometry is 1:1:1)

The concentrations of H₂A and HA⁻ will be the same, so Ka1 = [H⁺], and

pH = -log[H⁺]

6.70 = -log[H⁺]

[H⁺] =

[H⁺] = 2.00x10⁻⁷ M

Ka1 = 2.00x10⁻⁷

After the addition of 50.0 mL of NaOH the second equilibrium must dominate the reaction. For the expression of Ka1:

Ka1 = ([HA⁻]*[H⁺]/[H₂A])

[HA⁻]*[H⁺] = Ka1*[H₂A]

[HA⁻] = (Ka1*[H₂A])/[H⁺]

Ka2 = ([A⁻]*[H⁺]/[HA⁻])

Ka2 = ([A⁻]*[H⁺]²)/(Ka1*[H₂A])

By the stoichiometry, [H₂A] = [A⁻], so:

Ka2 = [H⁺]²/Ka1

pH = -log[H⁺]

8.00 = -log[H⁺]

[H+] = 10⁻⁸

Ka2 = (10⁻⁸)²/(2.00x10⁻⁷)

Ka2 = 5.00x10⁻¹⁰

hope this

explanation:

covalent bonds

consist of pairs of electrons shared by two atoms, and bind the atoms to fixed orientation and requre high energies to break them

ionic bonds

when two atoms differ considerably in their electronegativity, then one of the atoms will lose its electron to the other atom. this results in a positively charged ion(cation) and a negatively charged ion (anion)

explanation: no but i do not usually comment