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