Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of pH 6. If ionisation constant of HA is `10^(-5)`, the ratio of salt to acid concentration in the buffer solution will be :

Options

A. (a) `10:1`
B. (b) `4:5`
C. (c) `5:4`
D. (d) `1:4`

Answer

To solve the problem of finding the ratio of salt to acid concentration in a buffer solution formed by the addition of sodium hydroxide to a weak acid (HA), we can follow these steps:

### Step 1: Understand the Buffer Solution
When sodium hydroxide (NaOH) is added to a weak acid (HA), it reacts to form its conjugate base (A⁻), resulting in a buffer solution. The pH of this buffer solution is given as 6.

### Step 2: Use the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of the weak acid and the ratio of the concentrations of the salt (A⁻) to the acid (HA):
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]

### Step 3: Calculate pKa
The ionization constant (Ka) of the weak acid HA is given as \(10^{-5}\). We can calculate pKa using the formula:
\[ \text{pKa} = -\log(\text{Ka}) = -\log(10^{-5}) = 5 \]

### Step 4: Substitute Known Values into the Equation
Now, we can substitute the known values into the Henderson-Hasselbalch equation:
\[ 6 = 5 + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]

### Step 5: Solve for the Ratio of Salt to Acid Concentration
Rearranging the equation gives:
\[ 6 - 5 = \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
\[ 1 = \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]

To eliminate the logarithm, we can exponentiate both sides:
\[ 10^1 = \frac{[\text{A}^-]}{[\text{HA}]} \]
\[ \frac{[\text{A}^-]}{[\text{HA}]} = 10 \]

### Step 6: Express the Ratio
Thus, the ratio of salt (A⁻) to acid (HA) concentration is:
\[ \frac{[\text{A}^-]}{[\text{HA}]} = 10:1 \]

### Final Answer
The ratio of salt to acid concentration in the buffer solution is \(10:1\).

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