Suppose that 10.9 mL of HNO3 is neutralized by 83.9 mL of a 0.0021 < solution of KOH in a titration. Calculate the concentration of the HNO3 solution.
Answer in units of M
I assume the 0.0021 is M.
mols KOH = M x L = ?
mols HNO3 = mols KOH (from the coefficients in the equation.)
M HNO3 = mols HNO3/L HNO3. You know mols and L, solve for M.
To calculate the concentration of the HNO3 solution, we can use the concept of stoichiometry and the equation of the reaction between HNO3 and KOH.
The balanced equation for the reaction is:
HNO3 + KOH → KNO3 + H2O
Given that 10.9 mL of HNO3 is neutralized by 83.9 mL of a 0.0021 M solution of KOH, we need to determine the moles of KOH used in the reaction.
Step 1: Calculate the moles of KOH used.
Moles of KOH = Volume of KOH solution (in L) × Concentration of KOH (in M)
Moles of KOH = 83.9 mL × (1 L/1000 mL) × 0.0021 M
Moles of KOH = 0.00017619 mol
Step 2: Determine the moles of HNO3 present in the reaction.
Since the reaction is 1:1 between HNO3 and KOH, the moles of HNO3 used will be equal to the moles of KOH used.
Moles of HNO3 = 0.00017619 mol
Step 3: Calculate the concentration of the HNO3 solution.
Concentration of HNO3 = Moles of HNO3 / Volume of HNO3 solution (in L)
Concentration of HNO3 = 0.00017619 mol / (10.9 mL × (1 L/1000 mL))
Concentration of HNO3 = 0.01616 M
Therefore, the concentration of the HNO3 solution is 0.01616 M.