14. In deciding which of two acids is the stronger, one must know: A) the concen
ID: 229236 • Letter: 1
Question
14. In deciding which of two acids is the stronger, one must know: A) the concentration of each acid solution B) the pH of each acid solution C) the equilibrium constant of each acid D) all of the above E) both A and C. 15. Solid calcium hydroxide is dissolved in [OH 1 of the solution is water until the pt of the solution is 11.35. The hydroxide ion concentration A) 4.5× 10-12 M. B) 4.5× 103M C) 2.2× 10-3M. D) 1.1 x 103M E) none of these. 16. What volume of water must be added to 11.5 ml. of a pI 2.0 solution of HNO, in order to change the pH to 4.0 A) 11.5 mL B) 89 mL C) 114 mL D) 1.14 x 103 mL E) 29 ml 17he phi of HoCI is 7.5. Calculate the p of a 0.39 M solution of HOCL A) 7.50 B) 6.50 C) 3.95 D) 10.05 E) 0.39 3Explanation / Answer
14)
Answer
C) The equillibrium constant of each acid
Explanation
Consider an acid HA
HA <- - - - - > H+ + A-
the equillbrium constant expression is
Ka = [H+] [A-] /[HA]
15)
Answer
C) 2.2×10-3M
Explanation
pH + pOH = 14
pOH = 14 - pH
= 14 - 11.35
= 2.65
pOH = - log[OH-]
- log[OH-] = 2.65
[OH-] = 2.24×10-3M
16)
Answer
D) 1.14×10-3ml
Explanation
at pH = 2.0
[H+] = 10-2M
at pH = 4.0
[H+] = 10-4M
dilution of concentration = 10-2/10-4 = 100time
Total volume = 100×11.5ml = 1150ml
Volume of Water must be added = 1150-11.5 = 1.14×10-3ml
17)
Answer
C) 3.95
Explanation
pKa = - log Ka
Ka = 10-pKa = 10-7.5= 3.16×10-8
HClO <-----> H+ + ClO-
Ka = [H+] [ClO-] /[HClO]
at equillibrium
[HClO] = 0.39 - x
[ClO-] = x
[H+] = x
Therefore,
x2/0.39 - x = 3.16×10-8
we can assume, 0.39 - x ~ 0.39 because x is small value
x2 / 0.39 = 3.16×10-8
x2 = 1.23×10-8
x = 1.11×10-4
So,
[H+] = 1.11×10-4M
pH = - log(1.11×10-4) = 3.95
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