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A chemistry student weighs out 0.306 g of citric acid (H3C6H5O7) , a triprotic acid, into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits

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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

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