MP Board · Class 12 · Chemistry · Aldehydes, Ketones and Carboxylic AcidsA weak monocarboxylic acid (HA) has an acid dissociation constant ($Ka$) of $1.8 \times 10^{-5}$ at $298\text{ K}$. Calculate the hydrogen ion concentration ($H^+$), percent dissociation, and pH of a $0.05\text{ M}$ solution of this acid. Show all steps clearly.

To find the hydrogen ion concentration, percent dissociation, and pH of the weak monocarboxylic acid ($HA$), we use the dissociation equilibrium:

$$HA_{(aq)} \rightleftharpoons H^+{(aq)} + A^-{(aq)}$$

Given data:

• Concentration of acid ($C$) = $0.05\text{ M}$
• Dissociation constant ($K_a$) = $1.8 \times 10^{-5}$

Step 1: Calculate the hydrogen ion concentration ($[H^+]$)\nFor a weak acid, the degree of dissociation ($\alpha$) is given by: $$\alpha = \sqrt{\frac{K_a}{C}}$$ \nSubstituting the given values: $$\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{0.05}} = \sqrt{3.6 \times 10^{-4}} = 0.019$ \nSince $\alpha < 0.05$, our approximation is valid.\nThe hydrogen ion concentration $[H^+]$ is calculated as: $$[H^+] = C \times \alpha = 0.05 \times 0.019 = 9.5 \times 10^{-4}\text{ M}$$

Step 2: Calculate the percent dissociation $$\text{Percent dissociation} = \alpha \times 100$$ $$\text{Percent dissociation} = 0.019 \times 100 = 1.9%$$

Step 3: Calculate the pH $$\text{pH} = -\log[H^+]$$ $$\text{pH} = -\log(9.5 \times 10^{-4})$$ $$\text{pH} = -(\log(9.5) + \log(10^{-4}))$$ $$\text{pH} = -(\log(9.5) - 4) = 4 - \log(9.5)$$\nSince $\log(9.5) \approx 0.9777$: $$\text{pH} = 4 - 0.9777 = 3.0223$$

Final Answer:

• Hydrogen ion concentration $[H^+] = 9.5 \times 10^{-4}\text{ M}$
• Percent dissociation = $1.9%$
• pH = $3.02$