How much 0.10 M Base (in mL) is required to neutralize 8.00 mL of the 0.10 M acid?
Base Acid Base Volume (mL)
NaOH HCl
NaOH HC2H3O2
see below.
for HCL, which is a strong acid, the answer is 8.00 mL
For the other acid, which is a weak acid, you must use the ka formula
the answer is 7 for HCL and 8.72 for HC2H3O2
To solve this problem, you need to use the concept of molarity and the balanced chemical equation of the reaction between the base and the acid.
Let's consider the first case: NaOH (base) and HCl (acid).
1. Write the balanced chemical equation for the reaction:
NaOH + HCl → NaCl + H2O
2. Determine the number of moles of acid in the given volume (8.00 mL) using the formula:
Moles = Molarity × Volume (in liters)
Moles of HCl = 0.10 M × 8.00 mL ÷ 1000 mL/L = 0.008 moles (rounded to 3 decimal places)
3. From the balanced chemical equation, you can see that the stoichiometry between NaOH and HCl is 1:1. This means that they react in a 1:1 ratio.
Therefore, the number of moles of NaOH required to neutralize the HCl is also 0.008 moles.
4. Now, use the molarity and the amount of moles to find the volume of the base needed:
Volume (in mL) = Moles ÷ Molarity × 1000 mL/L
Volume of NaOH = 0.008 moles ÷ 0.10 M × 1000 mL/L = 80 mL
So, 80 mL of 0.10 M NaOH is required to neutralize 8.00 mL of 0.10 M HCl.
You can follow the same steps for the second case involving NaOH and HC2H3O2 to find the volume of NaOH required to neutralize it.