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

We are told that a weak monoprotic acid is represented as $$HA$$. When a solution of sodium hydroxide is added, some of the acid is neutralised, producing its conjugate base (the salt) $$A^-$$, and the resulting mixture behaves as an acid-base buffer.

The pH of this buffer is given as $$6$$ and the ionisation (dissociation) constant of the acid is $$K_a = 10^{-5}$$.

For any buffer that contains a weak acid $$HA$$ and its conjugate base (the salt) $$A^-$$ in appreciable amounts, we use the Henderson-Hasselbalch equation. First we state the formula:

$$\text{pH} = \text{p}K_a + \log\left(\dfrac{[\text{salt}]}{[\text{acid}]}\right)$$

Here, $$[\text{salt}]$$ stands for the molar concentration of the conjugate base $$A^-$$ (coming from the salt, sodium A), and $$[\text{acid}]$$ stands for the molar concentration of the undissociated weak acid $$HA$$ remaining in solution.

Now we substitute the given numerical data step by step. First, we convert the ionisation constant into its logarithmic form:

$$K_a = 10^{-5} \quad\Longrightarrow\quad \text{p}K_a = -\log K_a = -\log(10^{-5}) = 5$$

We also have the measured pH of the buffer:

$$\text{pH} = 6$$

Substituting $$\text{pH} = 6$$ and $$\text{p}K_a = 5$$ into the Henderson-Hasselbalch equation, we obtain:

$$6 = 5 + \log\left(\dfrac{[\text{salt}]}{[\text{acid}]}\right)$$

We next isolate the logarithmic term by subtracting $$5$$ from both sides of the equation:

$$6 - 5 = \log\left(\dfrac{[\text{salt}]}{[\text{acid}]}\right)$$

This simplifies to:

$$1 = \log\left(\dfrac{[\text{salt}]}{[\text{acid}]}\right)$$

To remove the logarithm, we use the fact that $$\log x = 1$$ implies $$x = 10^{1}$$, because the base of the common logarithm is $$10$$. Therefore:

$$\dfrac{[\text{salt}]}{[\text{acid}]} = 10^{1} = 10$$

In words, the concentration of the salt is ten times the concentration of the remaining acid. Expressed as a ratio, this is:

$$[\text{salt}] : [\text{acid}] = 10 : 1$$

Looking at the options provided, the ratio $$10 : 1$$ corresponds to Option A.

Hence, the correct answer is Option A.