1) Calculate the pH of a solution that contains 7.8 x 10-6 MOH A) 1.28 B) 5.11 C
ID: 1026332 • Letter: 1
Question
1) Calculate the pH of a solution that contains 7.8 x 10-6 MOH A) 1.28 B) 5.11 C) 12.72 D) 8.89 E) 9.64 2) Which of the following acids is the WEAKEST? The acid is followed by its Ka value. A) HC2H302, 1.8 × 10-5 B) HIO, 2.3 x 10-11 C) HBro, 2.3 × 10-9 D) HCIO, 2.9 x 10-8 BC6H5CO2H, 6.3 × 10-5 3) Determine the [H301 in a 0.265 M HCIO solution. The Ka of HCIO is 2.9 × 10-8. A) 1.1 × 10-10M B) 7.7 × 10-9 M C)1.3×10-6 M D) 4.9 x 10-4 M E) 8.8 x 10-5 M 4) Determine the pH of a 0.227 M C5H5N solution at 25°C. The Kb of C5H5N is 1.7 x 10-9 A) 4.59 B) 9.41 C) 4.71 D) 10.14 E) 9.29Explanation / Answer
1)
use:
pOH = -log [OH-]
= -log (7.8*10^-6)
= 5.1079
use:
PH = 14 - pOH
= 14 - 5.1079
= 8.89
Answer: D
2)
weakest acid will have smallest Ka value
The Ka is smallest for HIO
So, this is weakest
Answer: B
3)
HClO dissociates as:
HClO -----> H+ + ClO-
0.265 0 0
0.265-x x x
Ka = [H+][ClO-]/[HClO]
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 ((2.9*10^-8)*0.265) = 8.766*10^-5
since c is much greater than x, our assumption is correct
so, x = 8.766*10^-5 M
So, [H+] = x = 8.766*10^-5 M
Answer: E
4)
C5H5N dissociates as:
C5H5N +H2O -----> C5H5NH+ + OH-
0.227 0 0
0.227-x x x
Kb = [C5H5NH+][OH-]/[C5H5N]
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 ((1.7*10^-9)*0.227) = 1.964*10^-5
since c is much greater than x, our assumption is correct
so, x = 1.964*10^-5 M
So, [OH-] = x = 1.964*10^-5 M
use:
pOH = -log [OH-]
= -log (1.964*10^-5)
= 4.7068
use:
PH = 14 - pOH
= 14 - 4.7068
= 9.29
Answer: E
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