Answer:
1 M
Explanation:
The reaction equation is as follows:
CH3COOH + CH3COONa -------------------> 2CH3COO^- + NaH
One mole from each of the reactants yields 2 moles of acetate ions.
According to the problem, 2.00 mL, which is (2÷1000)L, of 0.50 M acetic acid reacts with 8.00 mL, equating to (8/1000)L, of 0.50 sodium acetate.
From the equation, we use n = CV -------------------------------------------(1).
Here, n = number of moles, V = volume, C = concentration.
The number of moles, n, of acetic acid = 0.50M × 2/1000L.
n(acetic acid) = 0.001 moles.
The number of moles, n, of sodium acetate = 0.50M × (8/1000)L.
n(sodium acetate) = 0.004 moles.
0.001 moles of acetic acid reacts with 0.004 moles of sodium acetate.
Thus, acetic acid acts as the limiting reagent.
One mole of acetic acid generates 2 moles of acetate ions.
0.001 mole of acetic acid results in = 0.002 moles of acetate ions.
According to the formula (1), n = CV.
0.002 = C × 2/1000
C = 0.002/0.002
C = 1 M
