Question

A flask contains 20.0mL of 0.10M sulfuric acid and 20.0mL of nitric acid. 50.0 mL of 0.20 M sodium hydroxide is required to neutralize the mixture of acids in the flask. What is the concentration of the nitric acid? I am confused on where to begin on this problem

Answer 1

find the moles of sodium hydroxide \((n_{NaOH})\)

find the moles of sulphuric acid \((n_{H_2SO_4})\)

subtract moles of sulphuric acid from moles of sodium hydroxide, the result is the moles of nitric acid \((n_{HNO_2})\). i.e. \(n_{HNO_2}=n_{NaOH}-n_{H_2SO_4}\)

find concentration of \(HNO_3\).

NOTE: account for the moles of the diprotic acid \((n_{H_2SO_4})\).

Answer 2

Thanks but I am confused because I got 3000, the answer is 0.30M, did I do a wrong conversion?

Answer 3

did you convert mL to L (because molarity is in units of "mol/L")?

Answer 4

yeah i had divided 60mol by 0.02L

Answer 5

did you multiply the moles of sulphuric acid by 2 because it's diprotic?

Answer 6

I did and still got 3000 by what i did above, i rounded it to 60mol

Answer 7

i don't think it's 60 mol, thats quite a lot.

moles of sulphuric acid: =(0.02 L*0.10M *2)
moles of sodium hydroxide = (0.05L* 0.20 M)

moles of nitric acid: (0.05L* 0.20 M)-(0.02 L*0.10M *2) =0.006 moles

Answer 8

oh I understand I actually figured out molar mass but now i understand thanks!

Answer 9

good stuff. no problem !