Question

A 0.115-L sample of an unknown HNO3 solution required 41.1 mL of 0.150 M Ba(OH)2 for complete neutralization. What was the concentration of the HNO3 solution?

Answer 1

Here for complete neutralization we need 41.1 mL of 0.15 M of

As we know the moles can be found as

Here for moles of , we have

Now the general complete balanced reaction is written as

Here it is observed that 2 moles of required to combine with 1 mole of

As we 1 mole of the molecule is equal to the molar mass of that molecule.

1 mole of = 63 g of

2 mole of = 126 g of

Now 1 mole of = 171.34 g of

Here we have 0.006165 moles of =

Now For the complete neutralization , we can say that

171.34 g of need 126 g of

1.06 g of will need the mass of as:

Now the concentration (molarity) of can obtained as by using the molarity equation

here w = given mass = 0.78 g

M = molar mass of = 63 g/mol

V= volume = 0.115 L = 115 mL

Now apply the molarity equation to calculate the concentration of

Hence the concentration of is