Need Help With Chem!!! (ch.8)?

2016-07-24 3:58 pm

Question 1:
Calculate the [OH−] of each aqueous solution with the following [H3O+].

A.) coffee, 1.2×10−5 M
B.) soap, 1.2×10−8 M
C.) cleanser, 5.1×10−10 M
D.) lemon juice, 2.5×10−2 M

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Question 2:
Calculate the [H3O+] of each aqueous solution with the following [OH−].

A.) vinegar, 1.2×10−11 M
B.) urine, 2.2×10−9 M
C.) ammonia, 5.7×10−3 M
D.) NaOH, 3.1×10−2 M

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Question 3:
Calculate the pH of each solution given the following [H3O+] or [OH−] values.
A.) [H3O+] = 6×10−4 M
B.) [H3O+] = 6×10−9 M
C.) [OH−] = 6×10−5 M
D.) [OH−] = 3.5×10−11 M
E.) [H3O+] = 6.3×10−8 M
F.) [OH−] = 8.3×10−4 M

回答 (1)

胡雪8°

2016-07-24 4:20 pm

✔ 最佳答案

Question 1:
A.)
[OH⁻] = Kw / [H₃O⁺] = (1.0 × 10⁻¹⁴) / (1.2 × 10⁻⁵) M = 8.3 × 10⁻¹⁰ M

B.)
[OH⁻] = Kw / [H₃O⁺] = (1.0 × 10⁻¹⁴) / (1.2 × 10⁻⁸) M = 8.3 × 10⁻⁷ M

C.)
[OH⁻] = Kw / [H₃O⁺] = (1.0 × 10⁻¹⁴) / (5.1 × 10⁻¹⁰) M = 2.0 × 10⁻⁵ M

D.)
[OH⁻] = Kw / [H₃O⁺] = (1.0 × 10⁻¹⁴) / (2.5 × 10⁻²) M = 4.0 × 10⁻¹³ M

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Question 2:
A.)
[H₃O⁺] = Kw / [OH⁻] = (1.0 × 10⁻¹⁴) / (1.2 × 10⁻¹¹) M = 8.3 × 10⁻⁴ M

B.)
[H₃O⁺] = Kw / [OH⁻] = (1.0 × 10⁻¹⁴) / (2.2 × 10⁻⁹) M = 4.5 × 10⁻⁶ M

C.)
[H₃O⁺] = Kw / [OH⁻] = (1.0 × 10⁻¹⁴) / (5.7× 10⁻³) M = 1.8 × 10⁻¹² M

D.
[H₃O⁺] = Kw / [OH⁻] = (1.0 × 10⁻¹⁴) / (3.1 × 10⁻²) M = 3.2 × 10⁻¹³ M

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Question 3:
A.)
pH = -log[H₃O⁺] = -log(6 × 10⁻⁴) = 3.2

B.)
pH = -log[H₃O⁺] = -log(6 × 10⁻⁹) = 8.2

C.)
pOH = -log[OH⁻] = -log(6 × 10⁻⁵) = 4.2
pH = 14.0 - pOH = 14.0 - 4.2 = 9.8

D.)
pOH = -log[OH⁻] = -log(3 × 10⁻¹¹) = 10.5
pH = 14.0 - pOH = 14.0 - 10.5 = 3.5

E.)
pH = -log[H₃O⁺] = -log(6.3 × 10⁻⁸) = 7.2

F.)
pOH = -log[OH⁻] = -log(8.3 × 10⁻⁴) = 3.1
pH = 14.0 - pOH = 14.0 - 3.1 = 10.9

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