Final answer:
The volume of HCl that is neutralized by the 7.0 M KOH solution is found by first calculating the moles of KOH used, then using the 1:1 molar ratio of the neutralization reaction to determine the moles of HCl neutralized, and finally converting this to the volume of a 4.0 M HCl solution, yielding a result of 371 mL.
Step-by-step explanation:
To find the volume of the HCl that is neutralized, we first use the stoichiometry of the neutralization reaction between HCl and KOH, which occurs in a 1:1 molar ratio:
HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)
Using the molarity (M) and volume (V) of the KOH solution, we can calculate the moles of KOH:
Moles of KOH = 7.0 M * 0.212 L = 1.484 mol
Since the reaction is 1:1, there are also 1.484 mol of HCl neutralized.
To find the volume of the 4.0 M HCl solution that contains this amount of HCl, use the formula for molarity (M = moles/V):
V(HCl) = Moles of HCl / Molarity of HCl
V(HCl) = 1.484 mol / 4.0 M = 0.371 L
Since 1 L = 1000 mL, converting to mL gives us:
V(HCl) = 0.371 L * 1000 mL/L = 371 mL
Therefore, the volume of HCl neutralized is 371 mL, which corresponds to the option (a).