Answer:
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