Question

A 0.100 m solution of a monoprotic weak acid has a ph of 3.00. what is the pka of this acid?

Answer 1

Step-by-step explanation:

It is known that the relation between pH and hydrogen ion concentration is as follows.

pH =
`-log [H^(+)]`

3 =
`-log [H^(+)]`

antilog (-3) =
`1 * 10^(-3)`

`[H^(+)] = 1 * 10^(-3)` M

Now, let the given acid is HA and it dissociates as follows.

`HA \rightarrow H^(+) + A^(-)`

0.1 0 0

0.1 - x x x

We know that relation between
`K_(a)`,
`[H^(+)]` and HA is as follows.

`K_(a) = ([H^(+)][A^(-)])/([HA])`

=
`(x * x)/((c - x))`

=
`(1 * 10^(-3) * 1 * 10^(-3))/((0.1 - 1 * 10^(-3)))`

=
`1.01 * 10^(-5)`

Also,
`pK_(a) = -log K_(a)`

=
`-log (1.01 * 10^(-5))`

= 5.00

Thus, we can conclude that the
`pK_(a)` of given acid is 5.0.

Answer 2

Answer is: pKa for the monoprotic acid is 5.
Chemical reaction: HA(aq) ⇄ A⁻(aq) + H⁺(aq).
c(monoprotic acid) = 0.100 M.
pH = 3.00.
[A⁻] = [H⁺] = 10∧(-3).
[A⁻] = [H⁺] = 0.001 M; equilibrium concentration.
[HA] = 0.1 M - 0.001 M.
[HA] = 0.099 M.
Ka = [A⁻]·[H⁺] / [HA].
Ka = (0.001 M)² / 0.099 M.
Ka = 0.00001 M = 1.0·10⁻⁵ M.
pKa = -logKa = 4.99.