Date Quiz 8-1 Acids and Bases pH
Name __________________________________________________ Period ________ Date ______________________ Quiz 8-1 Acids and Bases pH – Version 12
Total Points: _______/10
1. (a) Name the acid HClO4. Perchloric acid
(½ pt)
(b) Write the balanced dissociation equation for HClO4 in an aqueous solution. 𝐻𝐶𝑙𝑂4 (𝑎𝑞) → 𝐻 + (𝑎𝑞) + 𝐶𝑙𝑂4− (𝑎𝑞) (c) Determine the [H+] in 0.200 M HClO4.
(½ pt)
Answer and Units (½ pt)
Show Work (1/4 pt)
Sigfigs (1/4 pt)
Ans. & U (½ pt)
Work (1/4)
SF (1/4)
Ans. & U (½ pt)
Work (1/4)
SF (1/4)
Ans. & U (½ pt)
Work (1/4)
SF (1/4 pt)
𝐻𝐶𝑙𝑂4 (𝑎𝑞) → 𝐻 + (𝑎𝑞) + 𝐶𝑙𝑂4− (𝑎𝑞) 0.200 M 0.200 M 0.200 M [𝐻 + ] = 0.200 𝑀 (d) Calculate the pH of 0.200 M HClO4. 𝑝𝐻 = −𝑙𝑜𝑔[𝐻 + ] = −𝑙𝑜𝑔(0.200) = 0.699 (e) Calculate the pOH of 0.200 M HClO4. 𝑝𝐻 + 𝑝𝑂𝐻 = 14.00 𝑝𝑂𝐻 = 14.00 − 𝑝𝐻 = 14.00 − 0.699 = 13.301 (f) Calculate the [OH-] in 0.200 M HClO4. [𝑂𝐻 − ] = 10−𝑝𝑂𝐻 = 10−13.301 = 5.00 × 10−14 𝑀 2. (a) Write the chemical formula of rubidium hydroxide. RbOH
(½ pt)
(b) Write the balanced dissociation equation for rubidium hydroxide in an aqueous solution. 𝑅𝑏𝑂𝐻(𝑎𝑞) → 𝑅𝑏 + (𝑎𝑞) + 𝑂𝐻 − (𝑎𝑞)
(½ pt)
(g) Determine the [H+] in rubidium hydroxide pH 10.5 [𝐻 + ] = 10−𝑝𝐻 = 10−10.5 = 3 × 10−11 𝑀
Answer (½ pt)
Work (1/4)
SF (1/4 pt)
(h) Calculate the pOH of rubidium hydroxide pH 10.5 𝑝𝐻 + 𝑝𝑂𝐻 = 14.00
Answer (½ pt)
Work (1/4)
SF (1/4 pt)
Answer (½ pt)
Work (1/4)
SF (1/4 pt)
𝑝𝑂𝐻 = 14.00 − 𝑝𝐻 = 14.00 − 10.5 = 3.5 (i) Calculate the [OH-] in rubidium hydroxide pH 10.5 [𝑂𝐻 − ] = 10−𝑝𝑂𝐻 = 10−3.5 = 3 × 10−4 𝑀 3. When pH paper is added to HClO4 what color does it turn? red
4. When pH paper is added to a rubidium hydroxide solution, what color does it turn?
(½ pt)
(½ pt)
Name __________________________________________________ Period ________ Date ______________________ blue
Total Points: _______/10
1. (a) Name the acid HClO4. Perchloric acid
(½ pt)
(b) Write the balanced dissociation equation for HClO4 in an aqueous solution. 𝐻𝐶𝑙𝑂4 (𝑎𝑞) → 𝐻 + (𝑎𝑞) + 𝐶𝑙𝑂4− (𝑎𝑞) (c) Determine the [H+] in 0.200 M HClO4.
(½ pt)
Answer and Units (½ pt)
Show Work (1/4 pt)
Sigfigs (1/4 pt)
Ans. & U (½ pt)
Work (1/4)
SF (1/4)
Ans. & U (½ pt)
Work (1/4)
SF (1/4)
Ans. & U (½ pt)
Work (1/4)
SF (1/4 pt)
𝐻𝐶𝑙𝑂4 (𝑎𝑞) → 𝐻 + (𝑎𝑞) + 𝐶𝑙𝑂4− (𝑎𝑞) 0.200 M 0.200 M 0.200 M [𝐻 + ] = 0.200 𝑀 (d) Calculate the pH of 0.200 M HClO4. 𝑝𝐻 = −𝑙𝑜𝑔[𝐻 + ] = −𝑙𝑜𝑔(0.200) = 0.699 (e) Calculate the pOH of 0.200 M HClO4. 𝑝𝐻 + 𝑝𝑂𝐻 = 14.00 𝑝𝑂𝐻 = 14.00 − 𝑝𝐻 = 14.00 − 0.699 = 13.301 (f) Calculate the [OH-] in 0.200 M HClO4. [𝑂𝐻 − ] = 10−𝑝𝑂𝐻 = 10−13.301 = 5.00 × 10−14 𝑀 2. (a) Write the chemical formula of rubidium hydroxide. RbOH
(½ pt)
(b) Write the balanced dissociation equation for rubidium hydroxide in an aqueous solution. 𝑅𝑏𝑂𝐻(𝑎𝑞) → 𝑅𝑏 + (𝑎𝑞) + 𝑂𝐻 − (𝑎𝑞)
(½ pt)
(g) Determine the [H+] in rubidium hydroxide pH 10.5 [𝐻 + ] = 10−𝑝𝐻 = 10−10.5 = 3 × 10−11 𝑀
Answer (½ pt)
Work (1/4)
SF (1/4 pt)
(h) Calculate the pOH of rubidium hydroxide pH 10.5 𝑝𝐻 + 𝑝𝑂𝐻 = 14.00
Answer (½ pt)
Work (1/4)
SF (1/4 pt)
Answer (½ pt)
Work (1/4)
SF (1/4 pt)
𝑝𝑂𝐻 = 14.00 − 𝑝𝐻 = 14.00 − 10.5 = 3.5 (i) Calculate the [OH-] in rubidium hydroxide pH 10.5 [𝑂𝐻 − ] = 10−𝑝𝑂𝐻 = 10−3.5 = 3 × 10−4 𝑀 3. When pH paper is added to HClO4 what color does it turn? red
4. When pH paper is added to a rubidium hydroxide solution, what color does it turn?
(½ pt)
(½ pt)
Name __________________________________________________ Period ________ Date ______________________ blue
