2.)You are planning on titrating a 0.1200-g sample of malic acid (134.1 g/mol) w

Question

2.)You are planning on titrating a 0.1200-g sample of malic acid (134.1 g/mol) with 0.100-M NaOH. What volume of 0.100-M NaOH will be required to reach the first equivalence point?

A. 13.4 mL

B. 111.8 mL

C. 1.20 mL

D. 8.95 mL

3.)If it takes 10.00-mL to reach the first equivalence point in a titration, what volume will be required to reach the first half-equivalence point?

A. 5.00 mL

B. need more information to answer

C. 10.00 mL

4.)The first thing you will need to do is to prepare a ~0.1M solution of NaOH, which you will standardize. You will prepare 250-mL of this solution using a 30% by volume NaOH stock solution. How much NaOH will you need to prepare this solution? Report your answer to 3 significant digits with units of mL

5.)You calculate your 1/2 equivalence point to occur at 6.50-mL. You only collected data points at 6.00-mL and 8.00-mL. What is the pH at the 1/2 equivalence point? Volume (mL) pH 6.00 3.61 8.00 3.76

A. 3.96

B. 3.65

6.)You titrate 0.6895-g of an unknown acid with 0.250-M NaOH. It takes 16.50-mL of the NaOH to reach the first equivalence point. Given the following options, what is the identity of the unknown acid? A. quinolinic acid B. succinic acid C.malic acid

Explanation / Answer

Part 2)

Moles of malic acid = Mass/ Molar mass = 0.12/134.1 = 8.95 x 10-4

Moles of NaOH required at the equivalence point = C xV = 8.95 x 10-4

V = (8.95 x 10-4)/0.1 = 0.00895 L = 8.95 mL

Part 3)

Volume required to reach the first half-equivalence point = 10/2 = 5 mL

Part 4)

Molarity of 30 % NaOH solution = 7.5 M

Volume of 7.5 M NaOH required to prepare a 250 mL 0.1 M NaOH solution

=(250 x 0.1) / 7.5 = 3.33 mL

Part 5)

Here, the pH at the half way equivalence point at 6.5 mL must be 3.65

Part 6)

Moles of NaOH = 0.25M x 0.0165 L = 4.125 x 10-3

At the equivalence point, number of moles of unknown acid and NaOH becomes equal

Therefore, at equivalence point, Moles of unknown acid = 4.125 x 10-3

(Mass/Molar mass )=4.125 x 10-3

Molar mass = Mass/ 4.125 x 10-3 = 0.6895 g/ 4.125 x 10-3 = 167.15 g/mol

Molar mass of quinolinic acid = 167.12 g/mol

Therefore, unknown acid is quinolinic acid

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