Question

In: Chemistry

1. The Ka of a monoprotic weak acid is 6.11 × 10-3. What is the percent...

1. The Ka of a monoprotic weak acid is 6.11 × 10-3. What is the percent ionization of a 0.151 M solution of this acid? use quadratic equation to solve

2. A certain weak base has a Kb of 7.60 × 10-7. What concentration of this base will produce a pH of 10.23?

please show all your steps

thank you!

Solutions

Expert Solution

                                      HA ------> H+ + A-

initial                              c             0      0

change                       -c           c    c

at equilibrium           c(1-)       c       c

is a ionisation constant

Ka = [H+][A-]/[HA]

Ka    = c * c/c(1-)

Ka         = C2/1-    1<<<<<

Ka   = C2

6.11*10-3   = 0.1512

2   = 6.11*10-3 /0.151

= 4.046*10-2

= 0.2011

%   = 20.11%

2. PH = 10.23

POH = 14-PH

        = 14-10.23 = 3.77

POH = -log[OH-]

3.77   = -log[OH-]

[OH-] = 10-3.77 = 1.7*10-4 M

[OH-] = Kb*c

1.7*10-4   = 7.6*10-7 *C

(1.7*10-4)2   = 7.6*10-7 *C

C   = 0.038M

concentraction is 0.038M >>>> answer


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