Question

In: Chemistry

calculate the percent ionization of ha in a 0.10 m solution.

A certain weak acid, HA, has a Ka value of 6.7 X 10^-7.

Part A

Calculate the percent ionization of HA in a 0.10 M solution. Express your answer to two significant figures and include the appropriate units.

Part B

Calculate the percent ionization of HA in a 0.010 M solution.

Express your answer to two significant figures, and include the appropriate units.

Solutions

Expert Solution

Concepts and reason

This problem is based on the concept of ions and ionic equilibrium. A given substance on dissolution insolvent, which is generally water, yields either an ionic solution or a molecular solution. In an ionic solution substance splits up into ions, whereas in the latter, it is present.

Fundamentals

The extent of dissociation of a substance can be expressed in terms of the degree of dissociation, which is, by definition, equal to the fraction of total substance present in the form of ions.

 

(a) Consider a weak acid of type HA. It will dissociate as follows:

If \(\alpha\) is the degree of dissociation at a given concentration \(C\) of \(\mathrm{HA}\), then the concentrations in solution are as follows:

\([\mathrm{HA}]=C(1-\alpha)\)

\(\left[\mathrm{A}^{-}(a q)\right]=C \alpha\)

And, \(\left[\mathrm{H}^{+}(a q)\right]=C \alpha\)

since we know \(K_{a}\) is the dissociation constant and can be written as follows:

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

Substitute \(C(1-\alpha)\) for \([\mathrm{HA}], C \alpha\) for \(\left[\mathrm{A}^{-}\right]\) and \(\left[\mathrm{H}^{+}\right]\)

\(K_{a}=\frac{(C \alpha)(C \alpha)}{C(1-\alpha)}\)

\(=\frac{C^{2} \alpha^{2}}{C(1-\alpha)}\)

since \(\alpha\) is usually very small and thus negligible in comparison to unity. \((1-\alpha) \approx 1\)

Thus,

\(K_{a}=\frac{C \alpha^{2}}{1}\)

\(\alpha=\sqrt{\frac{K a}{C}}\)

Now substitute \(6.7 \times 10^{-7}\) for \(K_{a}\) and \(0.10 \mathrm{M}\) for \(C\) in the equation for \(\alpha\).

\(\alpha=\sqrt{\frac{6.7 \times 10^{-7}}{0.10}}\)

\(=0.00026\)

Percentage ionization is thus calculated as follows:

\(\%=\frac{(0.00026) 100}{0.10}\)

\(=0.26 \%\)

Here, \(\%\) is percentage ionization.

The value of the equilibrium constant is characteristic of a given weak electrolyte and depends only on the temperature. It is independent of the individual concentrations. Suppose a strong electrolyte is added to a weak electrolyte; even then, the same expression, as shown above, fits well. The effect of a strong electrolyte is to suppress the extent of dissociation of the weak electrolyte.

 

(b) The dissociation constant is written as follows:

$$ K_{a}=\frac{C \alpha^{2}}{1} $$

\(\alpha=\sqrt{\frac{K a}{C}}\)

Now substitute \(6.7 \times 10^{-7}\) for \(K_{a}\) and \(0.010 \mathrm{M}\) for \(C\) in the equation for \(\alpha\).

$$ \alpha=\sqrt{\frac{6.7 \times 10^{-7}}{0.010}} $$

\(=0.000081\)

Percentage ionization is thus calculated as:

\(\%=\frac{(0.000081) 100}{0.010}\)

\(=0.81 \%\)

Here, \(\%\) is percentage ionization.

 

The value of the equilibrium constant is characteristic of a given weak electrolyte and depends only on the temperature. It is independent of the individual concentrations. Suppose a strong electrolyte is added to a weak electrolyte; even then, the same expression, as shown above, fits well. The effect of a strong electrolyte is to suppress the extent of dissociation of the weak electrolyte.

 


Part a 

The percent ionization is \(0.26 \%\).

Part b

The percent ionization is \(0.81 \%\).

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