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

THE VOLUME OF NaOH NEEDED TO BE ADDED TO CITRIC ACID TO REACH THE EQUIVALENT POINT IS 4.725 L

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

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.