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 0.20 mol of sodium cyanate (NaCNO) and 1.0 mol of cyanic acid (HCNO) in enough water to make 1.0 litre of solution, we will follow these steps:

### Step 1: Identify the components of the buffer solution
The buffer solution consists of:
- A weak acid: HCNO (cyanic acid)
- Its conjugate base: NaCNO (sodium cyanate)

### Step 2: Determine the concentrations of the acid and the salt
Since we have 1.0 mol of HCNO and 0.20 mol of NaCNO in a total volume of 1.0 L:
- Concentration of HCNO = 1.0 mol / 1.0 L = 1.0 M
- Concentration of NaCNO = 0.20 mol / 1.0 L = 0.20 M

### Step 3: Calculate \( pK_a \)
Given \( K_a \) for HCNO is \( 2.0 \times 10^{-4} \):
- Calculate \( pK_a \) using the formula:
\[ pK_a = -\log(K_a) = -\log(2.0 \times 10^{-4}) \]
Using logarithmic properties:
\[ pK_a = -\log(2.0) - \log(10^{-4}) = -0.301 - (-4) = 4 - 0.301 = 3.699 \]

### Step 4: Use the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation for calculating the pH of a buffer solution is:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
Where:
- \([A^-]\) is the concentration of the conjugate base (NaCNO)
- \([HA]\) is the concentration of the weak acid (HCNO)

Substituting the values:
\[ pH = 3.699 + \log\left(\frac{0.20}{1.0}\right) \]
Calculating the logarithm:
\[ \log\left(\frac{0.20}{1.0}\right) = \log(0.20) = -0.699 \]
Now substituting back:
\[ pH = 3.699 - 0.699 = 3.000 \]

### Final Answer
The pH of the buffer solution is approximately **3.00**.

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