Calculate the concentration of acetate ion in a buffer solution made from 2.00 ml of 0.50 m acetic acid and 8.00 ml of 0.50 sodium acetate .

1 M

Explanation:

Equation of reaction is;

CH3COOH + CH3COONa > 2CH3COO^- + NaH

1 moles of each of the reactants react to give 2 moles of the acetate ion.

From the question, we have that 2.00 mL that is (2÷1000)L of 0.50 M acetic acid reacted with 8.00 mL that is (8/1000)L of 0.50 sodium acetate.

Then from equation, n = CV (1).

Where n= number of moles, V= volume, C= concentration.

Number of moles,n of acetic acid = 0.50M× 2/1000L.

n(acetic acid)= 0.001 moles.

Number of moles,n of sodium acetate= 0.50M ×(8/1000)L.

n(sodium acetate)= 0.004 moles.

0.001 moles of acetic acid react with 0.004 moles of Sodium acetate

Therefore, acetic acid is the limiting reagent.

One mole of acetic acid produces 2 moles of acetate ion.

0.001 mole of acetic acid produces= 0.002 moles of acetate ion.

Using the equation (1) that is, n= CV.

0.002= C× 2/1000

C= 0.002/0.002

C= 1 M