Titration Gizmo Explore Learning PDF

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11 Questions & Answers 1. What is the pH of an HCl solution if [H+ ] = 0.03? A. 0.03 B. 1.5 C. 3.0 D. 3.5

Correct Answer: B. 1.5 Explanation: To calculate the pH of a solution, Wnd the negative base-10 logarithm of the [H +] concentration:

pH = − log[ H+ ] − log[0.03] = 1.5

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2. The reaction of perchloric acid (HClO 4) with lithium hydroxide (LiOH) is described by the equation: HClO 4 + LiOH → LiClO 4 + H 2O Suppose 100 mL of perchloric acid is neutralized by exactly 46.9 mL of 0.75 M lithium hydroxide. What is the concentration of the perchloric acid? A. 0.35 M B. 0.47 M C. 0.63 M D. 1.60 M

Correct Answer: A. 0.35 M Explanation: Because 1 mole of perchloric acid reacts with 1 mole of lithium hydroxide, the moles of perchloric acid and lithium hydroxide at the equivalence point are equal. The moles of a substance can be found by multiplying its concentration by its volume:

moles HClO 4 = moles LiOH [HClO 4] ⋅ VHClO 4 = [LiOH] ⋅ VLiOH Solving for the concentration of perchloric acid:

[HClO 4] =

[LiOH]⋅V LiOH V [HClO4]

[HClO 4] =

0.75 M ⋅ 46.9 mL 100 mL

= 0.35 M

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3. The reaction of sulfuric acid (H2 SO4 ) with potassium hydroxide (KOH) is described by the equation: H2SO4 + 2KOH → K 2SO 4 + 2H 2O Suppose 50 mL of KOH with unknown concentration is placed in a iask with bromthymol blue indicator. A solution of 0.10 M H2 SO4 is dripped into the KOH solution. After exactly 33.4 mL of H2SO4 is added, the indicator changes from blue to yellow. What is the concentration of the KOH? A. 0.067 M B. 0.13 M C. 0.15 M D. 0.26 M

Correct Answer: B. 0.13 M

Explanation: Because 1 mole of sulfuric acid reacts with 2 moles of potassium hydroxide, there are twice as many moles of KOH at the equivalence point as moles of H2SO4:

moles KOH = 2 ⋅ moles H2 SO 4 [KOH] ⋅ VKOH = 2 ⋅ [H 2 SO4] ⋅ V H 2SO4 Solving for the concentration of potassium hydroxide:

[KOH] = [KOH] =

2⋅[H2SO 4 ]⋅V H

2 SO4

V KOH 2 ⋅ 0.1 M ⋅ 33.4 mL 50 mL

= 0.13 M

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4. The reaction of hydrochloric acid (HCl) with ammonia (NH3) is described by the equation: HCl + NH 3 → NH 4Cl A student is titrating 50 mL of 0.32 M NH3 with 0.5 M HCl. How much hydrochloric acid must be added to react completely with the ammonia? A. 6.4 mL B. 16.0 mL C. 32.0 mL D. 50.0 mL

Correct Answer: C. 32.0 mL Explanation: Because 1 mole of hydrochloric acid reacts with 1 mole of ammonia, the moles of hydrochloric acid and ammonia at the equivalence point are equal. The moles of a substance can be found by multiplying its concentration by its volume:

moles HCl = moles NH3 [HCl] ⋅ VHCl = [NH 3 ] ⋅ VNH3 Solving for the volume of hydrochloric acid:

VHCl = VHCl =

[ NH3 ]⋅VNH

3

[HCl] 0.32 M ⋅ 50 mL 0.5 M

= 32 mL

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5. Which type of titration is shown by the titration curve below?

A. Titration of a strong acid by a strong base B. Titration of a weak acid by a strong base C. Titration of a strong base by a strong acid D. Titration of a weak base by a strong acid

Correct Answer: D. Titration of a weak base by a strong acid

Explanation: The curve shows that the initial pH of the analyte is between 11 and 12 before the titrant is added. This indicates that the analyte is a base and the titrant is an acid. The equivalence point has a pH between 4 and 5. This low pH indicates that the base is a weak base. The product of a weak base and a strong acid is a salt that acts as an acid. This lowers the pH at the equivalence point below 7.0.

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