Calculate the pH of buffer solution prepared by dissolving 0.20 mol of sodium cyanate (NaCNO) and 1.0 mol of cynic acid (HCNO)in enough water to make 1.0 litre of solution . `K_a(HCNO)=2.0 xx10^(-4)`

Options

A. 0
B. `3.0`
C. `4.4`
D. `5.0`

Answer

To calculate the pH of the buffer solution prepared by dissolving sodium cyanate (NaCNO) and cyanic acid (HCNO), we will use the Henderson-Hasselbalch equation:

\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]

### Step 1: Determine the concentration of the salt and acid

1. **Concentration of Salt (NaCNO)**:
- Moles of NaCNO = 0.20 mol
- Volume of solution = 1 L
- Concentration of NaCNO = \(\frac{0.20 \text{ mol}}{1 \text{ L}} = 0.20 \text{ M}\)

2. **Concentration of Acid (HCNO)**:
- Moles of HCNO = 1.0 mol
- Volume of solution = 1 L
- Concentration of HCNO = \(\frac{1.0 \text{ mol}}{1 \text{ L}} = 1.0 \text{ M}\)

### Step 2: Calculate pKa from Ka

Given \( K_a(HCNO) = 2.0 \times 10^{-4} \):

\[ \text{pKa} = -\log(K_a) = -\log(2.0 \times 10^{-4}) \]

Using the logarithmic properties:

\[ \text{pKa} = -(\log(2.0) + \log(10^{-4})) = -\log(2.0) + 4 \]

Calculating \(\log(2.0)\):

\(\log(2.0) \approx 0.301\)

Thus,

\[ \text{pKa} = -0.301 + 4 = 3.699 \]

### Step 3: Substitute values into the Henderson-Hasselbalch equation

Now we can substitute the values into the Henderson-Hasselbalch equation:

\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]

Substituting the known values:

\[ \text{pH} = 3.699 + \log \left( \frac{0.20}{1.0} \right) \]

Calculating the logarithm:

\[ \log \left( \frac{0.20}{1.0} \right) = \log(0.20) = -0.699 \]

### Step 4: Final calculation of pH

Now substituting back into the pH equation:

\[ \text{pH} = 3.699 - 0.699 = 3.000 \]

### Conclusion

Thus, the pH of the buffer solution is:

\[ \text{pH} \approx 3.00 \]