When 0.076 moles of lead(II) nitrate are dis-
solved in enough water to make 670 mililiters
of solution, what is the molar concentration
of nitrate ions?
Answer in units of M
What's wrong with using M = moles/L soln? That will be M = 0.076/0.670 = ?
That is the molarity of lead(II)nitrate I need to find it for nitrate ions
There are two nitrate ions in one molecule of Pb(NO3)2; therefore, M Pb(NO3)2 x 2 = M NO3^-
To find the molar concentration of nitrate ions, we first need to calculate the number of moles of nitrate ions present in the lead(II) nitrate solution.
The chemical formula for lead(II) nitrate is Pb(NO3)2, which indicates that for every 1 mole of lead(II) nitrate, there are 2 moles of nitrate ions (NO3-).
Given that the solution contains 0.076 moles of lead(II) nitrate, we can determine the number of moles of nitrate ions by multiplying this value by the stoichiometric coefficient of nitrate ions in the compound, which is 2:
Number of moles of nitrate ions = 0.076 moles of Pb(NO3)2 × 2 moles of NO3-/1 mole of Pb(NO3)2
Number of moles of nitrate ions = 0.152 moles of NO3-
Now, to find the molar concentration (also known as molarity), we need to divide the number of moles of nitrate ions by the volume of the solution in liters. The given volume is 670 milliliters, which is equivalent to 0.670 liters:
Molar concentration of nitrate ions (M) = Number of moles of NO3- / Volume of solution (in liters)
Molar concentration of nitrate ions = 0.152 moles of NO3- / 0.670 liters
Molar concentration of nitrate ions = 0.227 M
Therefore, the molar concentration of nitrate ions in the solution is 0.227 M.