Q. how to calculate percent dissociation

Q: What is percent dissociation?

Percent dissociation is the fraction of substance that ionizes, expressed as percent. If initial concentration is c and dissociated amount is x, then percent dissociation =100\frac{x}{c}\%.

Q: What is the general formulfor percent dissociation?

Use percent dissociation =100\frac{x}{c}\%, where x is the equilibrium concentration of dissociated species and c is the initial concentration.

Q: How do I calculate percent dissociation for weak acid given K_a and initial concentration c?

Write the equilibrium expression \[K_a=\frac{x^2}{c-x}\] Solve for x, then percent =100\frac{x}{c}\%. Solve the quadratic when necessary.

Q: When can I use the small‑x approximation and how?

If x\ll c, approximate c-x\approx c. Then x\approx\sqrt{K_c}. Percent dissociation ≈ 100\sqrt{\frac{K_a}{c}}\%. Check that x/c<0.05 to validate the approximation.

Q: How do I set up an ICE table to find percent dissociation?

ICE: Initial: [HA]=c, [H^+]=0, [A^-]=0. Change: [HA]=c-x, [H^+]=x, [A^-]=x. Equilibrium: plug into K_a=\frac{x^2}{c-x}. Solve for x. Percent = 100\frac{x}{c}\%.

Q: How do I find K_a from known percent dissociation?

If percent dissociation = p\%, let \alpha=p/100. Then x=\alphc. Compute \[K_a=\frac{x^2}{c-x}=\frac{\alpha^2 c}{1-\alpha}.\]

Q: How is percent dissociation related to pH and pKa?

Using Henderson‑Hasselbalch, \frac{[A^-]}{[HA]}=10^{\mathrm{pH}-\mathrm{p}K_a}. Thus percent dissociation =100\frac{10^{\mathrm{pH}-\mathrm{p}K_a}}{1+10^{\mathrm{pH}-\mathrm{p}K_a}}\%.

Q: How do concentration and temperature affect percent dissociation?

For weak acids/bases, percent dissociation increases as concentration decreases, since \alpha\propto1/\sqrt{c}. Temperature changes percent according to the reaction enthalpy; endothermic dissociation increases with temperature, exothermic decreases.

Q: How do I handle bases or polyprotic acids?

For weak base use K_b analogously: set K_b=\frac{x^2}{c-x}. For polyprotic acids handle sequential dissociations with successive K_a values and appropriate equilibrium expressions for each step.

Answer

Percent dissociation is the fraction of an initial species that dissociates, expressed as a percent. For a weak acid HA with initial concentration \(C_0\) and acid dissociation constant \(K_a\), let \(x\) be the equilibrium \([H^+]\). Then
\[ K_a = \dfrac{x^2}{C_0 – x}. \]
This gives the quadratic
\[ x^2 + K_a x – K_a C_0 = 0, \]
so
\[ x = \dfrac{-K_a + \sqrt{K_a^2 + 4 K_a C_0}}{2}. \]
Percent dissociation is
\[ \text{Percent dissociation} = 100 \times \dfrac{x}{C_0} = 100 \times \dfrac{-K_a + \sqrt{K_a^2 + 4 K_a C_0}}{2C_0}. \]
For the common approximation \(x \ll C_0\), \(x \approx \sqrt{K_a C_0}\) and
\[ \text{Percent dissociation} \approx 100 \times \sqrt{\dfrac{K_a}{C_0}}.

Detailed Explanation

Definition. Percent dissociation is the fraction of an original weak electrolyte that separates into ions, expressed as a percentage. If a species HA dissociates into H+ and A-, then percent dissociation is

\[ \text{percent dissociation} = \frac{\text{amount dissociated}}{\text{initial amount}} \times 100\% \]

Step 1. Write the dissociation description in words. For a monoprotic weak acid HA: HA dissociates into H+ and A-. Define the initial concentration as \(C_0\). Let \(x\) be the concentration that dissociates at equilibrium. Then the equilibrium concentrations are \( [\mathrm{HA}] = C_0 – x\), \( [\mathrm{H}^+] = x\), and \( [\mathrm{A}^-] = x\).

Step 2. Write the equilibrium constant expression. For an acid with acid dissociation constant \(K_a\),

\[ K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]} = \frac{x \times x}{C_0 – x} = \frac{x^2}{C_0 – x} \]

Step 3. Solve for \(x\). Rearranging gives a quadratic in \(x\).

\[ x^2 + K_a x – K_a C_0 = 0 \]

Use the quadratic formula to obtain the exact solution for \(x\).

\[ x = \frac{-K_a + \sqrt{K_a^2 + 4 K_a C_0}}{2} \]

Choose the positive root because concentration must be positive. If \(x\) is small compared with \(C_0\) (a common case for weak acids), use the approximation \(C_0 – x \approx C_0\). Then

\[ K_a \approx \frac{x^2}{C_0} \quad\Rightarrow\quad x \approx \sqrt{K_a C_0} \]

Step 4. Calculate percent dissociation. Once \(x\) is known,

\[ \text{percent dissociation} = \frac{x}{C_0} \times 100\% \]

Approximate formula useful for quick estimates:

\[ \text{percent dissociation} \approx \sqrt{\frac{K_a}{C_0}} \times 100\% \]

Example. Find the percent dissociation of acetic acid with \(K_a = 1.8 \times 10^{-5}\) at initial concentration \(C_0 = 0.10\ \text{M}\).

Use the approximation \(x \approx \sqrt{K_a C_0}\). Compute

\[ x \approx \sqrt{(1.8 \times 10^{-5})(0.10)} = \sqrt{1.8 \times 10^{-6}} \approx 1.3416 \times 10^{-3}\ \text{M} \]

Percent dissociation:

\[ \text{percent dissociation} \approx \frac{1.3416 \times 10^{-3}}{0.10} \times 100\% \approx 1.34\% \]

Check the approximation by solving the quadratic exactly. Compute the discriminant and root:

\[ x = \frac{-K_a + \sqrt{K_a^2 + 4 K_a C_0}}{2} = \frac{-1.8 \times 10^{-5} + \sqrt{(1.8 \times 10^{-5})^2 + 4(1.8 \times 10^{-5})(0.10)}}{2} \]

Evaluating gives \(x \approx 1.333 \times 10^{-3}\ \text{M}\), and percent dissociation \( \approx 1.33\% \). The approximation was valid because the dissociation is much less than 5% of \(C_0\).

Notes and extensions. For a weak base, use \(K_b\) and the same algebra with \([\mathrm{OH}^-]\) in place of \([\mathrm{H}^+]\). For polyprotic acids or when ionic strength matters, set up the appropriate equilibrium expressions for each step and solve the coupled equations, or use numerical methods. Always check the small-\(x\) assumption by verifying that \(x/C_0\) is small (commonly < 5%).

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Chemistry FAQs

What is percent dissociation?

Percent dissociation is the fraction of substance that ionizes, expressed as percent. If initial concentration is \(c\) and dissociated amount is \(x\), then percent dissociation \(=100\frac{x}{c}\%\).

What is the general formulfor percent dissociation?

Use percent dissociation \(=100\frac{x}{c}\%\), where \(x\) is the equilibrium concentration of dissociated species and \(c\) is the initial concentration.

How do I calculate percent dissociation for weak acid given \(K_a\) and initial concentration \(c\)?

Write the equilibrium expression \[K_a=\frac{x^2}{c-x}\] Solve for \(x\), then percent \(=100\frac{x}{c}\%\). Solve the quadratic when necessary.

When can I use the small‑x approximation and how?

If \(x\ll c\), approximate \(c-x\approx c\). Then \(x\approx\sqrt{K_c}\). Percent dissociation ≈ \(100\sqrt{\frac{K_a}{c}}\%\). Check that \(x/c<0.05\) to validate the approximation.

How do I set up an ICE table to find percent dissociation?

ICE: Initial: \([HA]=c, [H^+]=0, [A^-]=0.\) Change: \([HA]=c-x, [H^+]=x, [A^-]=x.\) Equilibrium: plug into \(K_a=\frac{x^2}{c-x}\). Solve for \(x\). Percent = \(100\frac{x}{c}\%\).

How do I find \(K_a\) from known percent dissociation?

If percent dissociation = \(p\%\), let \(\alpha=p/100\). Then \(x=\alphc\). Compute \[K_a=\frac{x^2}{c-x}=\frac{\alpha^2 c}{1-\alpha}.\]

How is percent dissociation related to pH and pKa?

Using Henderson‑Hasselbalch, \(\frac{[A^-]}{[HA]}=10^{\mathrm{pH}-\mathrm{p}K_a}\). Thus percent dissociation \(=100\frac{10^{\mathrm{pH}-\mathrm{p}K_a}}{1+10^{\mathrm{pH}-\mathrm{p}K_a}}\%\).

How do concentration and temperature affect percent dissociation?

For weak acids/bases, percent dissociation increases as concentration decreases, since \(\alpha\propto1/\sqrt{c}\). Temperature changes percent according to the reaction enthalpy; endothermic dissociation increases with temperature, exothermic decreases.

How do I handle bases or polyprotic acids?

For weak base use \(K_b\) analogously: set \(K_b=\frac{x^2}{c-x}\). For polyprotic acids handle sequential dissociations with successive \(K_a\) values and appropriate equilibrium expressions for each step.

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