how many mL 2.3 M HCl is required to neutralize 85 g of KOH given the balanced reaction HCl + KOH → KCl + H2O? (molar masses: KOH = 56.1 g/mol; HCl = 36.45 g/mol)

0.66 mL are required for this neutralization

Explanation:

This is the neutralization equation

HCl + KOH → KCl + H₂O

When you neutralize, you make water like this:

HCl + H₂O → H₃O⁺ + Cl⁻

KOH → K⁺ + OH⁻

H₃O⁺ + OH⁻ ⇄ 2H₂O

Every time you neutralize, you have to apply this formula:

Nᵃ. Vᵃ = Nᵇ. Vᵇ

where N is normality and V volume

We also know that N. V = Gram equivalent

and N = M. valence

where in a base, valence means the quantity of OH inside the formula, and in an acid, valence means the quantity of H

In HCl, valence is 1 : in KOH is also 1

Gram equivalent is equal to mass in grams numerically equal to equivalent weight, where the equivalent weight (E) is the molar mass/valence

So 2.3. Vᵃ = Nᵇ. Vᵇ

Vᵃ is our unknown.

So for the base, the Nᵇ. Vᵇ = gram-equivalent

Gram equivalent = Mass / E

E = molar mass / valence

Gram equivalent KOH = 85g / (56.1 g/m / 1)

Gram equivalent KOH = 1.51

2.3. Vᵃ = 1.51

Vᵃ = 1.51 / 2.3 = 0.66 mL