I hope i\'m not asking for too much by asking for step by step solutions to the

Question

I hope i'm not asking for too much by asking for step by step solutions to the probelm, or atleast showing all your work. I want to understand this as thoroughly as possible because I am having a difficult time understanding how to do these problems, Thank you.

Practice Problem #1 1. Aspirin is a weak monoprotic acid. Calculate the volume of 0.493 M NaOH required to titrate a 0.1527-g sample of aspirin. (Pay attention to significant figures in your answer Part B. A student reached the end point of the titration in the previous question 0.34 mL before expected based on the volume calculated in Question 2. What is the purity of the sample? Part C. A student required more titrant than predicted as necessary in Question 2. Explain why this might occur

Explanation / Answer

Answer:

a) Molar mass of Aspirin = 180.157 g/mole.

Mass of Aspirin taken = 0.1527 g.

# of moles of Aspirin = Given mass / Molar mass = 0.1527/180.157 = 8.4759 x 10-4 moles.

When this solution is dissolved in aqueueous pahse the equal number of H+ ions will be there in solution (In presence of excess NaOH base there will be complete ionization of aspirin hence,

# of moles of H+ = # of moles of Aspirin = 8.4759 x 10-4 moles.

Let 'V' L of 0.493 M NaOH required for complete neutralization of Aspirin.'

# of moles of HO- = # of moles of NaOH = Molarity x Volume = 0.493V moles.

At neutralization,

# of moles of HO- =# of moles of H+ . ..............(1)

0.493V = 8.4759 x 10-4 moles

V = 8.4759 x 10-4 /0.493

V = 0.00172 L

V = 1.72 mL

Volume of NaOH required is 1.72 mL.

============================================

b) As neutralization take before addition of 0.34 mL of NaOH solution we can say that,

Actual volume of NaOH added = 1.72 - 0.34 = 1.38 mL.= 0.00138 L

With this volume of NaOH and using equation (1) let's calculate actual # of moles of Aspirin (i.e.H+)

# of moles Of Aspirin (H+) = Molarity x Actual volume of NaOH = 0.493 x 0.00138

# of moles Of Aspirin (H+) = 6.8 x 10-4 moles.

Then,

Actual Mass of Aspirin in sample = # of moles x Molar mass = 6.8 x 10-4 x 180.157 = 0.1226 g.

% Purity of Aspirin = (Actual Mass of Aspirin / Given mass of Aspirin sample) x 100

= (0.1226/0.1527) x 100

% Purity of Aspirin = 80.29 %

% purity of Aspirin is 80.29%.

=============================================

c) Question is clear enough.

========================XXXXXXXXXXXXX=====================

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---