What volume (in mL) of 0.45 M Na2SO4 solution is needed to precipitate all the barium, as BaSO4(s), from 14.5 mL of 0.15 M Ba(NO3)2 solution?
Ba(NO3)2(aq) + Na2SO4(aq)= BaSO4(s) + 2 NaNO3(aq)
mols Ba(NO3)2 = M x L = ?
mols Na2SO4 = mols Ba(NO3)2 (look at the coefficients in the balanced equation)
M Na2SO4 = mols Na2SO4/L Na2SO4. You now mols and M, solve for L and convert to mL.
To determine the volume of 0.45 M Na2SO4 solution needed to precipitate all the barium as BaSO4(s), we can use the stoichiometry of the balanced equation.
First, let's determine the stoichiometric ratio between Ba(NO3)2 and Na2SO4. Looking at the balanced equation:
Ba(NO3)2(aq) + Na2SO4(aq) = BaSO4(s) + 2 NaNO3(aq)
The ratio is 1:1 between Ba(NO3)2 and Na2SO4. This means that for every 1 mole of Ba(NO3)2, we need 1 mole of Na2SO4.
Now, let's calculate the moles of Ba(NO3)2 in the given solution. We can use the formula:
moles = concentration (M) × volume (L)
moles of Ba(NO3)2 = 0.15 M × 0.0145 L
moles of Ba(NO3)2 = 0.002175 mol
Since the ratio between Ba(NO3)2 and Na2SO4 is 1:1, we need 0.002175 mol of Na2SO4 to react with all the Ba(NO3)2.
Now, let's calculate the volume of 0.45 M Na2SO4 solution needed to provide 0.002175 mol of Na2SO4. We can rearrange the formula:
volume = moles / concentration
volume = 0.002175 mol / 0.45 M
volume ≈ 0.00483 L
Finally, to convert the volume to milliliters:
volume = 0.00483 L × 1000 mL/L
volume ≈ 4.83 mL
So, approximately 4.83 mL of 0.45 M Na2SO4 solution is needed to precipitate all the barium from 14.5 mL of 0.15 M Ba(NO3)2 solution.