The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL of 0.1 N phosphinic acid is __________.

We begin by recalling the fundamental principle for neutralisation reactions carried out in terms of equivalents. The number of equivalents of acid consumed must exactly equal the number of equivalents of base supplied. Mathematically, this is stated as

$$N_1 V_1 \;=\; N_2 V_2,$$

where $$N$$ represents the normality and $$V$$ the volume of the respective solutions. Here the subscripts 1 and 2 denote the acid and the base, respectively.

Now we identify each quantity from the data given in the problem:

• The phosphinic acid (the acid side) has a normality $$N_1 = 0.1\; \text{N}$$ and a volume $$V_1 = 10\; \text{mL}.$$

• The sodium hydroxide solution (the base side) has a normality $$N_2 = 0.1\; \text{N}$$, while its volume $$V_2$$ is what we have to determine.

Substituting these numerical values directly into the neutralisation equation, we get

$$0.1 \times 10 \;=\; 0.1 \times V_2.$$

To isolate $$V_2$$, we divide both sides of the equation by $$0.1$$:

$$V_2 \;=\; \dfrac{0.1 \times 10}{0.1}.$$

Simplifying the numerator, we first find $$0.1 \times 10 = 1.0$$, so the fraction becomes

$$V_2 \;=\; \dfrac{1.0}{0.1}.$$

Carrying out this final division, we obtain

$$V_2 = 10\; \text{mL}.$$

Hence the volume of $$0.1 \text{ N}$$ NaOH required to completely neutralise $$10 \text{ mL}$$ of $$0.1 \text{ N}$$ phosphinic acid is $$10 \text{ mL}$$.

So, the answer is $$10$$.