$$50\text{ mL}$$ of $$0.5\text{ M}$$ oxalic acid is needed to neutralize $$25\text{ mL}$$ of sodium hydroxide solution. The amount of $$\text{NaOH}$$ in $$500\text{ mL}$$ of the given sodium hydroxide solution is

Solution

Oxalic acid is dibasic, so the milliequivalents of acid are $$50 \times 0.5 \times 2 = 50$$. Then $$25 \times M = 50$$, so the $$\text{NaOH}$$ solution is $$2\text{ M}$$. $$500\text{ mL}$$ of it contains $$1$$ mol of $$\text{NaOH}$$, i.e. $$40\text{ g}$$.

Was this solution helpful?