1. Write the neutralization relation

For complete neutralization:

$\text{equivalents of acid} = \text{equivalents of base}$

If acid $A$ has basicity $n$, then 1 mole of acid gives $n$ acidic $H^+$ ions.

Base is $M(OH)_2$, so 1 mole of base gives 2 moles of $OH^-$. Hence its acidity factor is 2.

2. Calculate moles of acid

Given:

• Volume of acid $= 10\text{ mL} = 0.010\text{ L}$
• Molarity of acid $= 0.1\text{ M}$

So, moles of acid:

$n_A = 0.1 \times 0.010 = 0.001$

If its basicity is $x$, then acid equivalents are:

$\text{acid equivalents} = 0.001x$

3. Calculate moles of base

Given:

• Volume of base $= 30\text{ mL} = 0.030\text{ L}$
• Molarity of base $= 0.05\text{ M}$

So, moles of base:

$n_B = 0.05 \times 0.030 = 0.0015$

Each mole of $M(OH)_2$ gives 2 equivalents of $OH^-$. Therefore base equivalents are:

$\text{base equivalents} = 0.0015 \times 2 = 0.003$

4. Apply neutralization condition

$0.001x = 0.003$

$x = \frac{0.003}{0.001} = 3$

5. Final answer

The basicity of acid $A$ is:

$\boxed{3}$

6. Comparison with stored answer

Stored correct answer = 3

Our derived answer = 3

So, the derived answer agrees with the stored answer.