Question

Potable water (drinking water) should not have manganese concentrations in excess of 0.05 mg/mL. If the manganese concentration is greater than 0.1 mg/mL, it imparts a foul taste to the water and discolors laundry and porcelain surfaces. In an acidic solution, manganese(II) ion is oxidized to permanganate ion by bismuthate ion, BiO3-. In the reaction, BiO3- is reduced to Bi3+.

How many milligrams of NaBiO3 are needed to oxidize the manganese in 34.2 mg of manganese(II) sulfate?

Answer 1

14 H^+ + 2 Mn^2+ + 5 NaBiO3 --->5 Bi^3+ + 2 MnO4^- + 5 Na^+ + 7 H2O

Here is the the balanced reaction (Oxidation and Reduction)



34.2MgSO4=0.0342g of MgSO4

0.0342g of MgSO4*(1 mole/120.366 g)= moles of MgSO4


The equation shows that 2 MnSO4 moles=5 moles of NaBiO3

So,

moles of MgSO4*(5 moles of NaBiO3/ 2 moles of MnSO4)=moles of NaBiO3

moles of NaBiO3*(279.97 g/mol)=NaBiO3 in g

NaBiO3 in g*(10^3mg/g)= answer in mg


****Answer contains 3-significant figures