Answer:
THE VOLUME OF NaOH NEEDED TO BE ADDED TO CITRIC ACID TO REACH THE EQUIVALENT POINT IS 4.725 L
Step-by-step explanation:
The titration is between citric acid (H3C6H507) and NaOH
mass of citric acid = 0.306 g
Volume of citric acid = 250 mL = 250 /1000 = 0.25 L
Concentration of NaOH = 0.1000 M
Volume = unknown
First calculate the molar mass of citric acid
( 1 * 3 + 12* 6 + 1*5 + 16*7) = (4 + 72 + 5 + 112) = 193 g/mol
Since,
Concentration in moles/dm3 = concentration in g/dm3 / RMM
So the molarity of citric acid is:
Molarity = 0.306g / 0.25dm3 / Rmm
Molarity = 1.224g/dm3 / 193 g/mol
Molarity = 0.0063 M
Equation for the reaction is:
C3H5O(COOH)3 + 3NaOH → Na3C3H5O(COO)3 + 3H2O
Using the formula:
CaVa / CbVb = na/ nb
Ca = 0.0063 M
Cb = 0.1000 M
Va = 0.25 L
Vb = unknown
na = 1
nb = 3
Vb = Ca Va nb/ Cb na
Vb = 0.0063 * 0.25 * 3 / 0.1000 * 1
Vb = 0.4725 / 0.1000
Vb = 4.725 L
The volume of NaOH needed to reach the equivalent point is therefore 4.725 L