Question

7. What is the pH of a solution containing 5.3 x the pH of a solution containing 53 105 M hydroxide ion concentration a. 5.72 d.-4.27 b. 4.27 e.-9.72 C. 9.72 8. What is the H what is the pH of a solution containing o ly Macetic acid (K of acetic acid is 1.8 x 10") a. -0.77 d. 2.92 b. 1.98 e. 3.49 c. 2.22 -10). What is 9. One of the strongest acids known is called fluorosulfonic acid, HSO F (pK the poH of 1M of this acid? a. -0.053 d. 14 b. O e. 14.053 C. 9 10. Which of the following substances can be amphoteric? a. HSO4 d. H30* b. NH4* e. H3PO4 c. 02 11. For the following reaction, which is a conjugate acid and base pair? NH2CH3 (aq) + HCD2H (aq) + NH2CH3 (aq) + HCO3 (aq) a. Acid: NH2CH3; Conjugate Base: HCO2H b. Acid: HCOH; Conjugate Base: NH2CH3 C. Acid: NH2CH3; Conjugate Base: NH3CH3 d. Acid: HCO2H; Conjugate Base HCO2 e. Acid: NH3CH3; Conjugate Base: NH2CH3 12. If the K, of benzoic acid (CH3COOH) is 6.14 x 10-5, what is the pkn of the benzoate ion (C.HSCOO)? a. 1.63 x 10-10 d. 9.79 b. 6.14 x 10- e. Impossible to determine C. 4.21 13. What is the Ksp expression for the dissolving of Copper (II) arsenate (Cuz(AsO4)2)? a. Ksp = [Cu][AsO4] 3 d. Ksp = [Cut21[AsO4°12 b. Ksp = (Cu+2][AsO43] e. Ksp = [Cu 212 [AsO4"}} C. Ksp = [Cu] [AsO4]? 14. Which of these would be most likely to be weakly acidic? a. Ca(NO3)3 d. ki b. FeCl3 e. Mg(CIO)2 C. NaCl

Answer 1

7)

use:
pOH = -log [OH-]
= -log (5.3*10^-5)
= 4.2757


use:
PH = 14 - pOH
= 14 - 4.2757
= 9.7243
Answer: c

8)
CH3COOH dissociates as:

CH3COOH          ----->     H+   + CH3COO-
6*10^-3                 0         0
6*10^-3-x               x         x


Ka = [H+][CH3COO-]/[CH3COOH]
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 ((1.8*10^-5)*6*10^-3) = 3.286*10^-4

since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
1.8*10^-5 = x^2/(6*10^-3-x)
1.08*10^-7 - 1.8*10^-5 *x = x^2
x^2 + 1.8*10^-5 *x-1.08*10^-7 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.8*10^-5
c = -1.08*10^-7

Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 4.323*10^-7

roots are :
x = 3.198*10^-4 and x = -3.378*10^-4

since x can't be negative, the possible value of x is
x = 3.198*10^-4

So, [H+] = x = 3.198*10^-4 M


use:
pH = -log [H+]
= -log (3.198*10^-4)
= 3.49
Answer: e

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