Question

As part of Lab 11 you will make and standardize a solution of NaOH(aq). Suppose in the lab you measure the solid NaOH and dissolve it into 100.0 mL of water.

You then measure 0.1993 g of KHP (204.22 g/mol) and place it in a clean, dry 100-mL beaker, and then dissolve the KHP in about 25 mL of water and add a couple of drops of phenolphthalein indicator. You titrate this with your NaOH(aq) solution and find that the titration requires 9.92 mL of NaOH(aq).

Part 1:

How many moles of KHP are in your sample?

9.769×10^-4mol KHP

Ok.

Part 2:

How many moles of NaOH are required to react with this number of moles of KHP?

mol NaOH

Part 3:

What is the concentration of your NaOH(aq) solution?

M NaOH

Part 4:

You then take 0.1989 g of your aspirin sample, dissolve it in 20mL of ethanol, add a couple of drops of phenolphthalein, and titrate it with your NaOH(aq) solution. The titration requires 7.58 mL of NaOH(aq).

Using your calculated NaOH concentration, how many moles of NaOH were used in this titration of the aspirin?

mol NaOH

Part 5:

For this PreLab assignment we will assume all of the acid is the monoprotic acid acetylsalicylic acid (C₉H₈O₄). [Note: for your analysis in the PostLab you will subtract the moles of salicylic acid from the total acid content to determine the amount of acetylsalicylic acid].

How many moles of acetylsalicylic acid are in the 0.1989 g sample of your aspirin?

mol acid

Part 6:

How many grams of acetylsalicylic acid are in the 0.1989 g sample of your aspirin?

g

Part 7:

Determine the % by mass acetylsalicylic acid in your aspirin.

% acetylsalicylic acid

Answer 1

Part 2 : moles of NaOH = 9.769 x 10^-4mol

Part 3 : concentration of NaOH = 0.09838 M

Part 4 : moles of NaOH = 7.457 x 10^-4mol

Part 5 : moles acetylsalicylic acid = 7.457 x 10^-4mol

Part 6 : grams acetylsalicylic acid = 0.1343 g

Part 7 : % by mass acetylsalicylic acid = 67.54 %

Explanation

mass KHP = 0.1993 g

moles KHP = (mass KHP) / (molar mass KHP)

moles KHP = (0.1993 g) / (204.22 g/mol)

moles KHP = 9.759 x 10^-4 mol

moles NaOH = moles KHP

moles NaOH = 9.759 x 10^-4 mol

concentration NaOH = (moles NaOH) / (volume of NaOH used in Liter)

concentration NaOH = (9.759 x 10^-4 mol) / (9.92 x 10^-3 L)

concentration NaOH = 0.09838 M