Question

A 4.0 M solution of HCl is neutralized by 212 ml of a 7.0 M KOH solution. What is the volume of HCl that is neutralized?

a) 371 ml
b) 7.57 ml
c) 121 ml
d) 212 ml"

Answer 1

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).