20ml of 2M KOH are placed into a beaker. 4M CH 3 COOH are placed into a burette.

a. Predict what volume of CH 3 COOH will be needed to neutralize the KOH.

Respuesta :

Volume of CH3COOH will be needed to neutralize the KOH is 10 ml.

Explanation:

In the titration process the concentration of unknown analyte be it acid or base is calculated by the formula:

Macid x Vacid = Mbase x Vbase       equation 1

Data given:

volume of base KOH = 20 ml

molarity of base = 2M

Molarity of CH3COOH acid = 4 M

Volume  of CH3COOH acid = ?

Putting the values in the equation  1

4 x V acid = 20 x 2

V acid = [tex]\frac{40}{4}[/tex]

V acid = 10 ml

The neutralization of KOH by CH3COOH will be done by using 10 ml of 4M solution of CH3COOH against 20 ml of 2M KOH.

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