Question

0.460 grams of a diprotic acid are titrated to the endpoint with 31.5 mLof 0.200 N NaOH. What is the equivalent weight of the acid? What is the formula weight of the acid?

Answer 1

for NaOH, normality and molarity is same since NaOH has only 1 OH-

we have the Balanced chemical equation as:

2 NaOH + H2A ---> Na2A + 2 H2O

lets calculate the mol of NaOH

volume , V = 31.5 mL

= 3.15*10^-2 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.2*0.0315

= 6.3*10^-3 mol

From balanced chemical reaction, we see that

when 2 mol of NaOH reacts, 1 mol of H2A reacts

mol of H2A reacted = (1/2)* moles of NaOH

= (1/2)*6.3*10^-3

= 3.15*10^-3 mol

This is number of moles of H2A

mass of H2A = 0.46 g

we have below equation to be used:

number of mol = mass / molar mass

3.15*10^-3 mol = (0.46 g)/molar mass

molar mass = 146 g/mol

Since the acid is diprotic,

equivalent weight = molar mass/2

= 146/2

= 73 g/eq

equivalent weight = 73 g

formula weight = 146 g