What volume (in mL) of 0.55 M Na2SO4 solution is needed to precipitate all the barium, as BaSO4(s), from 15.5 mL of 0.25 M Ba(NO3)2 solution?
Ba(NO3)2 + Na2SO4 ==> BaSO4 + 2NaNO3
mols Ba(NO3)2 = M x L = ?
Using the coefficients in the balanced equation, convert mols Ba(NO3)2 to mols Na2SO4.
Then M Na2SO4 = mols Na2SO4/L Na2SO4. You know mols and M, solve for L and convert to mL.
To find the volume of the Na2SO4 solution needed, we need to determine the stoichiometry of the reaction between Ba(NO3)2 and Na2SO4. Based on the balanced equation, we know that 1 mole of Ba(NO3)2 reacts with 1 mole of Na2SO4 to produce 1 mole of BaSO4.
Step 1: Write down the balanced chemical equation for the reaction:
Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq)
Step 2: Calculate the moles of Ba(NO3)2 using the given concentration and volume:
moles of Ba(NO3)2 = concentration * volume
moles of Ba(NO3)2 = 0.25 M * 15.5 mL
moles of Ba(NO3)2 = 0.25 * 0.0155 mol
Step 3: Since the stoichiometry of the reaction is 1:1, the number of moles of BaSO4 formed is equal to the number of moles of Ba(NO3)2 used.
Step 4: Calculate the volume of the Na2SO4 solution needed:
moles of Na2SO4 = moles of BaSO4
concentration of Na2SO4 = moles of Na2SO4 / volume of Na2SO4
We can rearrange the equation to solve for the volume:
volume of Na2SO4 = moles of Na2SO4 / concentration of Na2SO4
Substituting the values:
volume of Na2SO4 = 0.25 * 0.0155 mol / 0.55 M
Simplifying,
volume of Na2SO4 = 0.004275 mol / 0.55 M
Solving for the volume of Na2SO4,
volume of Na2SO4 = 7.773 mL
Therefore, approximately 7.773 mL of the 0.55 M Na2SO4 solution is needed to precipitate all the barium as BaSO4 from 15.5 mL of 0.25 M Ba(NO3)2 solution.