Question

8) The Kh for A A) 9.33 is 4.9 10-10 What is the pH of a 0.0627 M aqueous NaA solution at 25.0 °C? B) 1.20 C) 8.75 D) 5.25 E)
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Answer #1

8)

A- dissociates as:

A- +H2O -----> HA + OH-

6.27*10^-2 0 0

6.27*10^-2-x x x

Kb = [HA][OH-]/[A-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((4.9*10^-10)*6.27*10^-2) = 5.543*10^-6

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

so, x = 5.543*10^-6 M

So, [OH-] = x = 5.543*10^-6 M

use:

pOH = -log [OH-]

= -log (5.543*10^-6)

= 5.25

use:

PH = 14 - pOH

= 14 - 5.25

= 8.75

Answer: 8.75

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