Pre-Lab Questions (Use a separate sheet of paper to answer the follouwing questi

Question

Pre-Lab Questions (Use a separate sheet of paper to answer the follouwing questions.) 1. Calcudlate the equivalen mass of ach of he fllowing dids. a. HC,H,O neutralize 0.5632 g of potassium hydrogen phthalate. phenolphthalein end point. Calculate the equivalent mass of the acid. b. KHCO, H,SO 2. Calculate the molarity of a solution of sodium hydroxide, NaOH, if 23.64 mL of this solution is needed to 3. It is found that 24.68 mL of 0.1165 M NaOH is needed to titrate 0.2931 g of an unknown acid to the 4. The following data was collected for the titration of 0.145 g of a weak acid with 0.100-M NaOH as the titrant: 0.00 5.00 10.00 12.50 15.00 20.00 24.00 2490 25.00 26.00 30.00 Volume NaOH added, mL pH | 2.88 4.15 4.58 476| 4.93| 536 6.14 715 .73| a. Use graph paper to graph the data. Place pH on the vertical axis and volume of NaOH on the horibontal 11.29 11.96 axis b. What is the pH at the equivalence point? c. Give the K, and pK, value of the acid. Explain.

Explanation / Answer

1.

(a)

CH3COOH (aq.) -----------> CH3COO- (aq.) + H+ (aq.)

Basicity = 1

Equivalent mass = molar mass / basicity of an acid = 60 / 1 = 60 equi.

(b)

KHCO3 (aq.) -----------> K+ (aq.) + HCO3- (aq.)

Equivalent mass = molar mass / valency = 100 / 1 = 100 equi.

(c)

H2SO3 -----------> 2 H+ (aq.) + SO32- (aq.)

Equivalent mass = molar mass / basicity = 82 / 2 = 41 equi.

2.

NaOH + KHT --------> NaKT + H2O

Moles of KHT = mass /molar mass = 0.5632 / 204.2 = 0.00276 mol

From the balanced equation,

1 mol KHT needs 1 mol NaOH

Then,

0.00276 mol of KHT needs 0.00276 mol of NaOH

Then,

Molarity of NaOH solution = moles of NaOH / Volume of solution in L = 0.00276 / 0.02364 = 0.1168 M

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