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 $$0.81 \%$$.