Calculate the volume in milliliters (ml) of a 2.48 m
kohsolution required to titrate (neutralize) 25.0 ml of 6.52 m
hclsolution.

65.72 mL

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = 6.52 M

V₁ = 25.0mL

M₂ = 2.48 M

V₂ = ?

Using the above formula , the volume of base , can be calculated as ,

M₁V₁ = M₂V₂

6.52 M  * 25.0mL  = 2.48 M * V₂

V₂ = 65.72 mL

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.

wher is the rest of the question