Buffers. Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (NaCNO) in enough water to make 1.0 liter of solution. [Ka(HCNO) = 2.0 × 10–4]
The Correct Answer and Explanation is:
To calculate the pH of a buffer solution composed of a weak acid (cyanic acid, HCNO) and its conjugate base (cyanate ion, CNO⁻, from sodium cyanate), we use the Henderson–Hasselbalch equation: pH=pKa+log([A−][HA])\text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
Step 1: Determine given values
- Moles of HCNO (acid) = 0.20 mol
- Moles of NaCNO (conjugate base) = 0.80 mol
- Total volume = 1.0 L
- Therefore, concentrations:
- [HCNO]=0.201.0=0.20 M[HCNO] = \frac{0.20}{1.0} = 0.20\, \text{M}
- [CNO−]=0.801.0=0.80 M[CNO^-] = \frac{0.80}{1.0} = 0.80\, \text{M}
Step 2: Calculate pKa\text{p}K_a
Ka=2.0×10−4⇒pKa=−log(2.0×10−4)≈3.70K_a = 2.0 \times 10^{-4} \Rightarrow \text{p}K_a = -\log(2.0 \times 10^{-4}) \approx 3.70
Step 3: Use the Henderson–Hasselbalch equation
pH=3.70+log(0.800.20)\text{pH} = 3.70 + \log \left( \frac{0.80}{0.20} \right) pH=3.70+log(4)\text{pH} = 3.70 + \log(4) pH=3.70+0.60=4.30\text{pH} = 3.70 + 0.60 = 4.30
✅ Final Answer: pH = 4.30
Explanation
Buffer solutions are critical in maintaining stable pH levels in chemical and biological systems. A buffer consists of a weak acid and its conjugate base (or a weak base and its conjugate acid). When a small amount of acid or base is added to a buffer, it resists significant pH change due to the equilibrium between the acid and base components.
In this example, cyanic acid (HCNO) is a weak acid, and sodium cyanate (NaCNO) is its conjugate base in salt form. When NaCNO dissolves in water, it dissociates completely, releasing cyanate ions (CNO⁻). These ions can react with any added H⁺, while the HCNO can donate H⁺ to neutralize added OH⁻—providing buffering action.
The Henderson–Hasselbalch equation is a convenient tool for calculating the pH of such buffer systems. It relates the pH to the acid dissociation constant (KaK_a) and the ratio of conjugate base to acid concentrations.
Here, with 0.80 mol of CNO⁻ and 0.20 mol of HCNO, the ratio [CNO−][HCNO]\frac{[\text{CNO}^-]}{[\text{HCNO}]} is 4. Taking the log of 4 gives 0.60. The KaK_a value provided for HCNO allows us to find its pKa\text{p}K_a (3.70), and when we plug values into the Henderson–Hasselbalch equation, the resulting pH is 4.30.
This pH value reflects a slightly acidic buffer (as expected, since HCNO is a weak acid), and the relatively high amount of conjugate base helps resist drops in pH, making this an effective buffer system.
