Concentration of OH- left in a solution made by mixing 16.0 g of NaOH and 80.0 mL of 0.170M HNO3?
HNO3 + NaOH ==> NaNO3 + H2O
moles HNO3 = M x L = 0.0136
moles NaOH = g/molar mass = 16/40 = 0.4
So all of the HNO3 will be used up and NaOH remaining will be 0.400 - 0.0136 = ??
and molarity = moles/L = ??
To find the concentration of OH- left in the solution, we need to determine the limiting reactant in the reaction between NaOH and HNO3, and then use stoichiometry to calculate the concentration of OH-.
Step 1: Determine the limiting reactant
First, we need to find the number of moles of NaOH and HNO3 present in the solution.
Mass of NaOH = 16.0 g
Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.99 g/mol
Number of moles of NaOH = mass / molar mass = 16.0 g / 39.99 g/mol = 0.40 mol
Volume of HNO3 = 80.0 mL = 0.080 L
Concentration of HNO3 = 0.170 M (moles per liter)
Number of moles of HNO3 = concentration × volume = 0.170 mol/L × 0.080 L = 0.0136 mol
From the balanced chemical equation:
NaOH + HNO3 → NaNO3 + H2O
We can see that the stoichiometry of the reaction is 1:1 for NaOH and HNO3. Therefore, the limiting reactant is the one with the smaller number of moles, which is HNO3.
Step 2: Calculate the concentration of OH-
Since HNO3 is the limiting reactant, it will react completely with NaOH in a 1:1 ratio. This means that all of the moles of HNO3 react with the same number of moles of NaOH.
The number of moles of OH- that will be produced is also equal to the number of moles of HNO3 (since the stoichiometry is 1:1), which is 0.0136 mol.
Now, we need to calculate the final volume of the solution. Since NaOH is a strong base, it dissociates completely in water, so the total volume of the solution will include the volume of HNO3 and any water formed from the reaction.
The reaction between NaOH and HNO3 produces water, so the total volume of the final solution is equal to the sum of the initial volume of HNO3 and the volume of water formed.
Volume of HNO3 = 80.0 mL = 0.080 L
Volume of water formed = number of moles of HNO3 × molar volume of water = 0.0136 mol × 22.4 L/mol = 0.3054 L
Total volume of the final solution = Volume of HNO3 + Volume of water formed = 0.080 L + 0.3054 L = 0.3854 L
Finally, we can calculate the concentration of OH- left in the solution. Since the moles of OH- is equal to the moles of HNO3, we can use the formula:
Concentration of OH- = moles of OH- / total volume of solution
Concentration of OH- = 0.0136 mol / 0.3854 L = 0.0353 M
Therefore, the concentration of OH- left in the solution is 0.0353 M.