14 In deciding which of two acids is the stronger one must k

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 3

Solution

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