A 0.110-L sample of an unknown HNO3 solution required 52.1 mL of 0.150 M Ba(OH)2 for complete neutralization. What was the concentration of the HNO3 solution?
2HNO3 + Ba(OH)2 ==> 2H2O + Ba(NO3)2
mols Ba(OH)2 = M x L = ?
Use the equation coefficients to convert mols Ba(OH)2 to mols HNO3. Note: that's mols HNO3 = 2*mols Ba(OH)2.
Then M HNO3 = mols HNO3/L HNO3. You know L and mols, solve for M.
To find the concentration of the HNO3 solution, we can use the concept of stoichiometry.
1. Write the balanced equation for the reaction between HNO3 and Ba(OH)2:
HNO3 + Ba(OH)2 → Ba(NO3)2 + H2O
2. Determine the moles of Ba(OH)2 used in the reaction:
Moles of Ba(OH)2 = volume of Ba(OH)2 solution (in L) × molarity of Ba(OH)2 solution
Moles of Ba(OH)2 = 52.1 mL × (0.150 mol/L / 1000 mL/L) = 0.007815 mol
3. Based on the balanced equation, we know that 1 mole of Ba(OH)2 reacts with 2 moles of HNO3. Therefore, the moles of HNO3 used in the reaction are twice the moles of Ba(OH)2 used:
Moles of HNO3 = 2 × Moles of Ba(OH)2 = 2 × 0.007815 mol = 0.01563 mol
4. Finally, calculate the concentration of the HNO3 solution:
Concentration of HNO3 = Moles of HNO3 / Volume of HNO3 solution
Concentration of HNO3 = 0.01563 mol / 0.110 L = 0.142 M
Therefore, the concentration of the HNO3 solution is 0.142 M.
Are either of these right?
52.1 mL of .150M solution contains
.0521*.150 = .00782 moles of Ba(OH)2
Since the reaction is
2HNO3 + Ba(OH)2 = Ba(NO3)2 + 2H2O
we need .0156 moles of HNO3
.0156moles/.110L = 0.142moles/L = .142M