12.b Due: Thursday 11/16/2017 (Last Day of class) e. None of the above Part 2: S

Question

12.b

Due: Thursday 11/16/2017 (Last Day of class) e. None of the above Part 2: Short Answer/Calculations (Answer all questions in this section) 2. Calculate pH, pOH and the percent fraction of dissociation species for each of the following solutions. See the table provided for Ka information. a. 0.00000075 moles of Lithium Hydroxide (LiOH) in 2L of water (6 pts) b. /2.50 g lactic acid (HC HsO) in 250 ml. of water (6 pts) 0.67 M Periodate (104) (6 pts) (6 pts) d.) 0.0085 M Nitric Acid (HNO3) 5IP age

Explanation / Answer

12. Calculations

a. molarity of LiOH solution = moles/L

                                             = 0.00000075 mol/2 L = 3.75 x 10^-7 M

LiOH is a strong base dissociates comletely into Li+ and OH-

So,

[OH-] = 3.75 x 10^-7 M

pOH = -log[OH-] = -log(3.75 x 10^-7) = 6.426

pH = 14 - pOH = 7.574

Percent dissociation = 100%

b. molarity of lactic acid solution = 2.50 g/90.08 g/mol x 0.250 L

                                                    = 0.111 M

HC3H5O3 + H2O <==> C3H5O3- + H3O+

let x amount disscoiated

Ka = [C3H5O3-][H3O+]/[HC3H5O2]

1.37 x 10^-4 = x^2/0.111

x = [H3O+] = 3.90 x 10^-3 M

pH = -log[H3O+] = 2.41

pOH = 14 - pH = 11.59

percent dissociation = (3.90 x 10^-3/0.111) x 100 = 3.51%

c. periodate = 0.67 M

It is an conjugate base of strong acid HIO4

IO4- + H2O <==> HIO4 + OH-

Kb = 1 x 10^-14/2.3 = x^2/0.67

x = [OH-] = 5.4 x 10^-8 M

pOH = -log[OH-] = 7.27

pH = 14 - pOH = 6.73

percent dissociation = 5.4 x 10^-8 x 100/0.67 = 8.06 x 10^-6%

d. HNO3 = 0.0085 M

HNO3 is strong acid completely dissociates into H+ and NO3-

[H+] = 0.0085 M

pH = -log[H+] = 2.07

pOH = 14 - pH = 11.93

percent dissociation = 100%

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