Final answer:
To neutralize 20.0 mL of 0.100 M HCl, 10.0 mL of 0.200 M NaOH is required. The calculations are based on the 1:1 mole ratio of the neutralization reaction between HCl and NaOH and using the molarity and volume of both the acid and base.
Step-by-step explanation:
The question involves a neutralization reaction in which hydrochloric acid (HCl) is titrated with sodium hydroxide (NaOH). The neutralization reaction can be represented by the equation:
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
This is a 1:1 reaction, meaning one mole of HCl reacts with one mole of NaOH. To determine the volume of NaOH needed to neutralize a given volume and concentration of HCl, use the formula:
Molarity of Acid (M1) × Volume of Acid (V1) = Molarity of Base (M2) × Volume of Base (V2)
Here are the given values:
- M1 (HCl) = 0.100 M
- V1 (HCl) = 20.0 mL
- M2 (NaOH) = 0.200 M
- V2 (NaOH) = Unknown
First, calculate the moles of HCl:
Moles of HCl = M1 × V1 = 0.100 M × 20.0 mL = 0.002 moles
Since the reaction ratio is 1:1, the moles of NaOH required will be the same:
Moles of NaOH = 0.002 moles
Now we solve for the volume of NaOH:
V2 = Moles of NaOH / M2 = 0.002 moles / 0.200 M = 0.010 L or 10.0 mL
Therefore, the volume of 0.200 M NaOH required to neutralize 20.0 mL of 0.100 M HCl is 10.0 mL.