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Question: 12.5 mL of a solution containing 6.0 g of a dibasic acid in 1 L water was found to be neutralized by...

12.5 mL of a solution containing 6.0 g of a dibasic acid in 1 L water was found to be neutralized by 10 mL of a decinormal solution of NaOH. The molecular weight of the acid (gmol−1)(gmol^{-1}) is:

A

150

B

120

C

110

D

75

Answer

150

Explanation

Solution

The mass of the dibasic acid in 1 L (1000 mL) of the solution is 6.0 g. The volume of the acid solution taken is 12.5 mL. Mass of acid in 12.5 mL solution = (6.0 g/1000 mL)×12.5 mL=0.075 g(6.0 \text{ g} / 1000 \text{ mL}) \times 12.5 \text{ mL} = 0.075 \text{ g}. The solution is neutralized by 10 mL of a decinormal (0.1 N) NaOH solution. Using the neutralization law, milli-equivalents of acid = milli-equivalents of base: Nacid×Vacid=Nbase×VbaseN_{acid} \times V_{acid} = N_{base} \times V_{base}. Nacid×12.5 mL=0.1 N×10 mLN_{acid} \times 12.5 \text{ mL} = 0.1 \text{ N} \times 10 \text{ mL}. Nacid=(0.1×10)/12.5=1/12.5=0.08 NN_{acid} = (0.1 \times 10) / 12.5 = 1 / 12.5 = 0.08 \text{ N}. The relationship between normality (N), molarity (M), and basicity (n-factor) of an acid is N=M×nN = M \times n. Since the acid is dibasic, its basicity (n-factor) is 2. 0.08 N=Macid×20.08 \text{ N} = M_{acid} \times 2. Macid=0.08/2=0.04 MM_{acid} = 0.08 / 2 = 0.04 \text{ M}. Molarity is defined as moles of solute per liter of solution. Macid=(Mass of acid/Molecular Weight)/Volume of solution (in L)M_{acid} = (\text{Mass of acid} / \text{Molecular Weight}) / \text{Volume of solution (in L)}. 0.04 mol/L=(0.075 g/MW)/(12.5/1000 L)0.04 \text{ mol/L} = (0.075 \text{ g} / \text{MW}) / (12.5 / 1000 \text{ L}). 0.04=0.075/(MW×0.0125)0.04 = 0.075 / (\text{MW} \times 0.0125). MW=0.075/(0.04×0.0125)\text{MW} = 0.075 / (0.04 \times 0.0125). MW=0.075/0.0005\text{MW} = 0.075 / 0.0005. MW=150 g/mol\text{MW} = 150 \text{ g/mol}.