Question

Percent Ionization

Percent ionization for a weak acid (HA) is determined by the following formula:

Percent ionization=[HA] ionized[HA] initial×100%

For strong acids, ionization is nearly complete (100%) at most concentrations. However, for weak acids, the percent ionization changes significantly with concentration. The more diluted the acid is, the greater percent ionization.
A certain weak acid, HA, has a Ka value of 7.6×10−7.
Part A
Calculate the percent ionization of HA in a 0.10 M solution.
Express your answer as a percent using two significant figures.

%

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Part B
Calculate the percent ionization of HA in a 0.010 M solution.
Express your answer as a percent using two significant figures.

%

Answer 1

(a) 0.26 % (b) 0.80 %

Step-by-step explanation:

(a)

Given that:

`K_(a)=7.6* 10^(-7)`

Concentration = 0.10 M

Considering the ICE table for the dissociation of acid as:-

The expression for dissociation constant of acid is:

`K_(a)=\frac {\left [ H^(+) \right ]\left [ {A}^- \right ]}{[HA]}`

`7.2* 10^(-7)=(x^2)/(0.10-x)`

`7.2\left(0.10-x\right)=10000000x^2`

Solving for x, we get:

x = 0.00026 M

Percentage ionization =
`(0.00026)/(0.10)* 100=0.26 \%`

(b)

Concentration = 0.010 M

Considering the ICE table for the dissociation of acid as:-

The expression for dissociation constant of acid is:

`K_(a)=\frac {\left [ H^(+) \right ]\left [ {A}^- \right ]}{[HA]}`

`7.2* 10^(-7)=(x^2)/(0.010-x)`

`7.2\left(0.010-x\right)=10000000x^2`

Solving for x, we get:

x = 0.00008 M

Percentage ionization =
`(0.00008)/(0.010)* 100=0.80 \%`