Calculate the hydronium and hydroxide concentration of a solution made by mixing 100 mL 0.020 mol/L NaOH and 25.0 mL of 0.040 mol/L HCl.
I'm having trouble knowing where to start with this question.
Another question very similar to the one stated above.
Suppose 1.00 x 10-2 mol of KOH and 1.0 x 10-4 mol of HCI are both added to 1.00 L of
water. Calculate the [H3O+] and [OH─] of the resulting solution.
#1.
mols HCl = M x L = ?
mols NaOH = M x L = ?
Subtract to find excess of which is present.
If excess HCl, that is the mols HCl.
If excess NaOH, that is the mols NaOH.
Then M (of the excess) = mols/total L. That gives either H3O^+ or OH-. The other one is found from
(H3O^+)(OH^-) = Kw = 1E-14.
#2 is done similarly except the problem has calculated mols already.
Wow DrBob222, thanks for the clear and concise explanation.
To solve this question, you need to know the concept of neutralization reactions and the stoichiometry of the reaction involved.
First, let's understand what happens when NaOH (a strong base) and HCl (a strong acid) react together. They neutralize each other to form water and a salt. The balanced chemical equation is:
NaOH + HCl -> H2O + NaCl
From the balanced equation, we can see that one mole of NaOH reacts with one mole of HCl to form one mole of water and one mole of NaCl.
To solve the problem, you can follow these steps:
Step 1: Calculate the number of moles of NaOH and HCl used.
We are given the concentrations and volumes of NaOH and HCl solutions. Use the formula:
moles = concentration (mol/L) x volume (L)
For NaOH:
moles of NaOH = 0.020 mol/L x 0.100 L = 0.002 mol
For HCl:
moles of HCl = 0.040 mol/L x 0.025 L = 0.001 mol
Step 2: Determine the limiting reagent.
From the balanced equation, we know that one mole of NaOH reacts with one mole of HCl. The reactant that has fewer moles is the limiting reagent, meaning it will be completely consumed during the reaction, and the other reactant will be in excess.
In this case, both NaOH and HCl have the same number of moles. So, either NaOH or HCl could be considered the limiting reagent.
Step 3: Calculate the number of moles of water formed.
Since NaOH and HCl react in a 1:1 mole ratio, the number of moles of water formed will be the same as the number of moles of NaOH or HCl used.
In this case, 0.002 moles of water will be formed.
Step 4: Calculate the volume of the final solution.
To calculate the volume of the final solution, add the volumes of the NaOH and HCl solutions.
Volume of final solution = Volume of NaOH + Volume of HCl
= 100 mL + 25 mL
= 125 mL = 0.125 L
Step 5: Calculate the concentration of the products.
Since the volume of the final solution is 0.125 L and we formed 0.002 moles of water, we can calculate the concentration of the products.
Concentration of H2O = moles of water / volume of final solution
= 0.002 mol / 0.125 L
= 0.016 mol/L
Since water is a neutral substance, the concentration of hydronium ions (H3O+) and hydroxide ions (OH-) is equal. Therefore, the concentration of H3O+ and OH- in the final solution is both 0.016 mol/L.
So, the hydronium and hydroxide concentration of the solution is 0.016 mol/L.