Question

1. In the second step she prepared the vinegar solution for the titration in the following manner 25.0 mL of vinegar were diluted to 250.0 mL in a volumetric flask, and 25.0 mL of this diluted solu- tion required 8.27 mL of the above standardized NaOH (gquestion 5a) to reach the phenolphthalein endpoint. What is the molarity of the acid as it was titrated (diluted vinegar)? a. Use the dilution equation to determine the molarity of the acid prior to dilution (i., the real molarity of the vinegar). b. What is the percentage by mass of the acid in the original vinegar sample? c.

Answer 1

Q5a)

Moles of KHP = Mass/Molar Mass= (1.092 g) / (204.23 g/mol) = 0.005347 mol

Volume of NaOH = 48.76 mL= 48.76 mL *(0.001 L/mL)= 0.04876 L

KHP is a monoprotic acid. Thus, moles of NaOH= moles of KHP = 0.005347 mol

Thus, we have,

Moles of NaOH = Molarity * Volume

Molarity of NaOH = Moles of NaOH/Volume = 0.005347 mol / 0.04876 L = 0.1096 M

Q1)

a)

Let molarity of 25.0 mL Vinegar be M_a.

Volume of Vinegar, V_a = 25.0 M

Volume of NaOH, V_b= 8.27 mL

Molarity of NaOH, M_b = 0.1096 M

Thus, we have,

Meva = M V MV Ma = V1 (0.1096M) (8.27mL) 25.0mL = 0.03627M

b)

0.03627 M is the molarity of 25.0 mL solution which was taken from 250 mL solution.

Thus, molarity of 250.0 mL diluted solution is the same that is 0.03627.

The dilution equation is

M,V= Mala

Here, M means molarity and V means Volume. The subscript 'u' is for undiluted and 'd' is for diluted solution.

Thus, we have,

M, 25.0mL = (0.03627M)(250.0mL) (0.03627M)(250.0mL) Mu= = 0.3627M 25,0ml

c)

Volumes in L of undiluted solution = 25.0 mL * (0.001L/mL)= 0.0250 L

Moles of Acetic Acid = M_uV_u = 0.3627 mol/L * 0.0250 L = 0.0090675 mol

Molar Mass of Acetic Acid = 60.052 g/mol

Mass of Acetic Acid = Moles*Molar Mass = 0.0090675 mol * 60.052 g/mol = 0.5446 g

Let density of Vinegar be 'd' g/ml

Thus, Mass of Vinegar solution = Volume * Density = 25.0 mL * d g/mL= 25.0d g

Percentage by mass

Mass of Acid % by mass = Mass of Vinegar - X 100 = 0.5446 00 x 100 = 2.18d%

Note that here, density needs to be given. The density d of vinegar is usually 1.05 g/mL

Assuming d= 1.05 g/mL, we will have weight % as 2.18 * 1.05 = 2.28%