A weak acid `HA` has `K_(a) = 10^(-6)`. What would be the molar ratio of this acid and its salt with strong base so that `pH` of the buffer solution is `5`?

Options

A. `1//10`
B. `10`
C. `1`
D. `2`

Explanation

To solve the problem step by step, we need to find the molar ratio of the weak acid \( HA \) and its salt \( A^- \) in a buffer solution that has a pH of 5. The dissociation constant \( K_a \) of the weak acid is given as \( 10^{-6} \).

### Step 1: Calculate \( pK_a \)
The first step is to calculate the \( pK_a \) from the given \( K_a \).

\[ pK_a = -\log(K_a) = -\log(10^{-6}) = 6 \]

### Step 2: Use the Henderson-Hasselbalch equation
Next, we use the Henderson-Hasselbalch equation, which is given by:

\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]

We know that \( pH = 5 \) and \( pK_a = 6 \). We can substitute these values into the equation:

\[ 5 = 6 + \log\left(\frac{[A^-]}{[HA]}\right) \]

### Step 3: Rearrange the equation
Now, we can rearrange the equation to isolate the logarithmic term:

\[ \log\left(\frac{[A^-]}{[HA]}\right) = 5 - 6 \]
\[ \log\left(\frac{[A^-]}{[HA]}\right) = -1 \]

### Step 4: Convert from logarithmic to exponential form
To eliminate the logarithm, we convert it to exponential form:

\[ \frac{[A^-]}{[HA]} = 10^{-1} = 0.1 \]

### Step 5: Find the molar ratio
Now, we can express the molar ratio of \( HA \) to \( A^- \):

\[ \frac{[HA]}{[A^-]} = \frac{1}{0.1} = 10 \]

### Conclusion
Thus, the molar ratio of the weak acid \( HA \) to its salt \( A^- \) is \( 10:1 \).

### Final Answer
The molar ratio of the weak acid \( HA \) and its salt \( A^- \) is \( 10:1 \).

---