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Pepsi [2]
3 years ago
9

What is the pH of 0.10 M NaF(aq). The Ka of HF is 6.8 x 10-4

Chemistry
1 answer:
dsp733 years ago
6 0

<u>Given information:</u>

Concentration of NaF = 0.10 M

Ka of HF = 6.8*10⁻⁴

<u>To determine:</u>

pH of 0.1 M NaF

<u>Explanation:</u>

NaF (aq) ↔ Na+ (aq) + F-(aq)

[Na+] = [F-] = 0.10 M

F- will then react with water in the solution as follows:

F- + H2O ↔ HF + OH-

Kb = [OH-][HF]/[F-]

Kw/Ka = [OH-][HF]/[F-]

At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x

10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x

x = [OH-] = 1.21*10⁻⁶ M

pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92

pH = 14 - pOH = 14-5.92 = 8.08

Ans: (b)

pH of 0.10 M NaF is 8.08

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A nitric acid solution flows at a constant rate of 5L/min into a large tank that initially held 200L of a 0.5% nitric acid solut
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Answer:

x(t) = −39e

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Explanation:

Let V (t) be the volume of solution (water and

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measured in liters after t minutes, and let c(t) be the concentration (by volume) of

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c(t) = x(t)

V (t)

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We model this problem as

dx

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both measured in liters of nitric acid per minute. The input rate is

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·

20 Lnit.

100 Lsol.

=

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The output rate is

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1 min

·

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dx

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dx

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1000 Lsol.

=

x(0) Lnit.

200 Lsol.

.

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In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is

µ(t) = exp Z

P(t) dt

= exp

0.03 Z

dt

= e

0.03t

.

The solution is

x(t) = 1

µ(t)

Z

µ(t)Q(t) dt + C

= Ce−0.03t + 1.2e

−0.03t

Z

e

0.03t

dt

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1.2

0.03

e

−0.03t

e

0.03t

= Ce−0.03t +

1.2

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The constant is found using x(t) = 1:

x(0) = Ce−0.03(0) + 40 = C + 40 = 1.

Thus C = −39, and the solution is

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