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

Solution

Chemistry ma’am · 4 steps

Both strengths are already in normality, so equal equivalents means equal volumes here.

1. Equivalents must match. At neutralisation, the gram-equivalents of acid equal the gram-equivalents of base. Normality times volume counts the equivalents (milliequivalents when the volume is in mL).

   N_(acid) V_(acid) = N_(base) V_(base)

2. Basicity is built in. Phosphinic acid, H₃PO₂, has only one P–OH hydrogen, so it is monobasic. The other two H atoms sit directly on P and are not acidic. Because the strength is given as normality, its basicity is already counted in the 0.1 N.
3. Put in numbers. 0.1 × 10 = 0.1 × V, which gives V = 10 mL.
4. So. The volume of 0.1 N NaOH needed is 10 mL.

The common slip: 30. Treating H₃PO₂ as tribasic and multiplying by 3 gives 30 mL. The acid is monobasic, and the strength is already in normality, so there is nothing to multiply.

The idea to keep. When both solutions are given in normality, use N₁V₁ = N₂V₂ directly. Basicity only matters when you convert molarity to normality.

Answer from the NTA final answer key. Solution by PrepWiser.