Question

A 30.0 mL sample of 2.3 M Cu(OH)2 sample was titrated to with 48.5 mL of HNO3 to find

Answer 1

Answer:

2.8 M HNO3

Explanation:

                 Cu(OH)2 + 2HNO3 ---> Cu(NO3)2 + 2H2O

                 1 mol           2 mol

The number of moles Cu(OH)2 in the sample:

2.3 mol/L*0.0300L = 0.069 mol

The number of moles HNO3  in the solution of HNO3 that was used for titration

x mol/L*0.0485L = 0.0485x mol

From equation we can see that number of moles HNO3 = 2 times number of moles Cu(OH)2.

So, we can write

0.0485x = 2*0.069

x = 2*0.069/0.0485 = 2.8 M HNO3