According to this, calculate the concentration of lead (II) nitrate (331.2 g/mol) as molarity and g/L?

Question

4. A solution of 25 mL Na2CO3 is neutralized with 22.4 mL of 0.1065 M HCl. In the another analysis, 50 mL of Na2CO3 solution is added to 25.0 mL of Pb(NO3)2 solution, and after the lead (II) carbonate was filtrated, the excess of Na2CO3 is neutralized with 14.8 mL of the same acid. According to this, calculate the concentration of lead (II) nitrate (331.2 g/mol) as molarity and g/L?

Answer 1

Hi Bilge, I just uploaded my video tutorial for this problem onto youtube and you can access it through this link: https://youtu.be/OZUYSXDeXCk

Hope this helps!

Kalvin