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Exercises · 6.60

Q.The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

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A weak acid partially ionizes; given pH, we find [H+][\text{H}^+] and use it to calculate both KaK_a (from the equilibrium expression) and α\alpha (the fraction ionized). For 0.1 M HCNO with pH 2.34: Ka=2.19×10−4K_a = 2.19 \times 10^{-4} and α=4.57%\alpha = 4.57\%.

Cyanic acid is a weak acid, meaning it doesn't fully dissociate in water. Instead, it establishes an equilibrium between the undissociated molecule and its ions. The pH tells us the hydrogen ion concentration at equilibrium, which is the key to unlocking both the ionization constant KaK_a (a measure of acid strength) and the degree of ionization α\alpha (the fraction of molecules that actually ionized).

The equilibrium we're dealing with is:

HCNO⇌H++CNO−\text{HCNO} \rightleftharpoons \text{H}^+ + \text{CNO}^-

The ionization constant is defined as:

Ka=[H+][CNO−][HCNO]K_a = \frac{[\text{H}^+][\text{CNO}^-]}{[\text{HCNO}]}

Let's work through this systematically.

Step-by-step solution

1. Extract the hydrogen ion concentration from pH

The definition of pH gives us:

pH=−log⁡[H+]\text{pH} = -\log[\text{H}^+]

So:

[H+]=10−pH=10−2.34=4.57×10−3 M[\text{H}^+] = 10^{-\text{pH}} = 10^{-2.34} = 4.57 \times 10^{-3} \text{ M}

This is the equilibrium concentration of hydrogen ions.

2. Recognize the stoichiometry at equilibrium

For every HCNO molecule that ionizes, we get one H+\text{H}^+ and one CNO−\text{CNO}^-. Since the acid is the only source of both ions:

[H+]=[CNO−]=4.57×10−3 M[\text{H}^+] = [\text{CNO}^-] = 4.57 \times 10^{-3} \text{ M}

3. Find the equilibrium concentration of undissociated acid

We started with 0.1 M HCNO. Some of it ionized (the amount that ionized equals [H+][\text{H}^+]), so:

[HCNO]equilibrium=0.1−4.57×10−3=0.09543 M[\text{HCNO}]_{\text{equilibrium}} = 0.1 - 4.57 \times 10^{-3} = 0.09543 \text{ M}

Tip

For very weak acids where [H+]≪c0[\text{H}^+] \ll c_0, you can approximate [HCNO]≈c0[\text{HCNO}] \approx c_0. Here, 4.57×10−34.57 \times 10^{-3} is about 4.6% of 0.1, so the approximation would introduce a small error but is often acceptable for quick estimates.

4. Calculate the ionization constant KaK_a

Substitute the equilibrium concentrations into the expression:

Ka=[H+][CNO−][HCNO]=(4.57×10−3)(4.57×10−3)0.09543K_a = \frac{[\text{H}^+][\text{CNO}^-]}{[\text{HCNO}]} = \frac{(4.57 \times 10^{-3})(4.57 \times 10^{-3})}{0.09543}

Ka=2.088×10−50.09543=2.19×10−4K_a = \frac{2.088 \times 10^{-5}}{0.09543} = 2.19 \times 10^{-4}

So Ka=2.19×10−4K_a = 2.19 \times 10^{-4} (using the exact equilibrium concentration 0.1−0.00457=0.09540.1 - 0.00457 = 0.0954 M; the quick approximation with c≈0.1c \approx 0.1 gives 2.09×10−42.09 \times 10^{-4}). …

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