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1. Calculate the pH of the following solutions. Show all of your work. a. 0.0050 M nitric acid, HNO3 b. 0.50 M hydrocyanic ac

2. A student was given the following problem: Calculate the pH of a 0.50 Mama bromide (NHaBr) solution if the Kb of NH3 is 1.

3. A student was given the following problem: Calculate the pH of a 0.250 M potassium acetate (KCH3CO2) solution if the Ka of

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Answer #1

1)

a)

[H+] = [HNO3]

= 0.0050 M

use:

pH = -log [H+]

pH = -log (0.0050)

= -log (5.0*10^-3)

= 2.301

Answer: 2.30

b)

HCN dissociates as:

HCN -----> H+ + CN-

0.5 0 0

0.5-x x x

Ka = [H+][CN-]/[HCN]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((4*10^-10)*0.5) = 1.414*10^-5

since c is much greater than x, our assumption is correct

so, x = 1.414*10^-5 M

So, [H+] = x = 1.414*10^-5 M

use:

pH = -log [H+]

= -log (1.414*10^-5)

= 4.8495

Answer: 4.85

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