I\'m stuck on this, please show working out please A chemist needs to determine

Question

I'm stuck on this, please show working out please

A chemist needs to determine the concentration of a solution of nitric acid, HNO3. She puts 815mL of the acid in a flask along with a few drops of indicator. She then slowly adds 0.400M Ba(OH)2 to the flask until the solution turns pink, indicating the equivalence point of the titration. She notes that 105mLof Ba(OH)2 was needed to reach the equivalence point.

Solution map

In this titration, the concentration of base is known and can be used to calculate the unknown acid concentration:

concentration of base ? moles of base ? moles of acid ? concentration of acid

Part A

How many moles of Ba(OH)2 are present in 105mL of 0.400M Ba(OH)2?

Express your answer with the appropriate units.

Explanation / Answer

Solution:

Given Volume of HNO3 = 815 mL (convert to L )

Molarity of Ba(OH)2 = 0.400 M , Volume of Ba(OH)2 = 105 mL (convert to L )

Volume of HNO3 in L = 815 mL * 1 L / 1000 mL = 0.815 L

Volume of Ba(OH)2 in L = 105 mL * 1 L / 1000 mL = 0.105 L

Mole of Ba(OH)2 = Molarity* volume in L = 0.400 M * 0.105 L = 0.042

Lets write the reaction between HNO3 and Ba(OH)2

2 HNO3 + Ba(OH)2   -> 2H2O + BaNO3

The mole ratio of HNO3 to Ba(OH)2 is 2 : 1

We use mol of Ba(OH)2 to get moles of HNO3

Mol HNO3 = mol Ba(OH)2 *2 mol HNO3 / 1mo Ba(OH)2

= 0.042 mol Ba(OH)2 * 3 mol HNO3/ 1 mol Ba(OH)2

=0.084 mol HNO3

Now concentration of HNO3 = mol HNO3 / volume in L = 0.084 mol HNO3 / 0.815 L

=0.103 M

[HNO3]= 0.103 M

Part A

How many moles of Ba(OH)2 are present in 105mL of 0.400M Ba(OH)2?

Express your answer with the appropriate units.

Mole of Ba(OH)2 = Molarity* volume in L = 0.400 M * 0.105 L = 0.042

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects