Question

Figure out the pH of solutions A, C, and D.

Solution A is 40.0 mL of .100 M weak acid HX.

Solution B is 0.500 M salt NaX and has a pH of 10.02

Solution C is made by adding 10.0 mL of 0.250 M KOH to solution A.

Solution D is made by adding 10.0 mL of 0.100 M HCl to Solution C.

Each pH is rounded to 2 decimal places. Hint: start with solution B

in regards to the "expert" who commented, the ka is not needed if you know how to do this question properly. The kb needs to be found from solution B first and then used for A because the X is the same.

Answer 1

Solution B :    pH = 10.02 , pOH = 14-10.02 = 3.98 , [OH^-] = 10^-pOH = 10^-3.98 = 0.000105   

   X^- + H₂O -------> HX    + OH^-

initial    0.5    0    0

   Change    -0.000105    + 0.000105 +0.000105

Equilibrium 0.499895 0.000105    0.000105

Kb = [OH^-] [HX]/[X^-] = (0.000105)²/0.499895 = 1.1025 x 10^-8/0.499895 = 2.205 x 10^-8 pKb = -log Kb = -log (0.00000002205) = 7.66

Ka = Kw/kb = 10^-14/2.205 x 10^-8 = 0.45 x 10^-6 , pKa = -log Ka = -log (0.00000045) = 6.35

Solution A :    HX + H₂O ---> X^- + H₃O⁺    HX = molarity x vol in lts = 0.1 M x 40/1000 = 0.004

   Initial moles 0.004    0    0

   Change -x +x    +x

Equilibrium 0.004-x    x x

Ka = [H₃O⁺] [X^-] [HX] = x²/0.004-x

X is insignificant in 0.004-x

X²/0.004 = 0.45 x 10^-6

x² = 0.45 x 10^-6 x 0.004 = 0.0018 x 10^-6

x = 4.24 x 10^-5 M

pH = -log [H₃O⁺] = - log (4.24 x 10^-5) = 4.37

Solution C : HX + OH^- ---> X^- + H₂O    Moles of KOH = M x V = 0.25 M x 10/1000 = 0.0025

   Initial 0.004 0.0025 0 0   

   Change    -0.0025 -0.0025 + 0.0025

   Equilibrium 0.0015    0 0.0025

According to Henderson - Hasselbalch equation :

pH = pKa + log [X^-]/[HX]

pH = 6.35 + log (0.0025/0.0015)

pH = 6.35 + log 1.67

pH = 6.35 + 0.22 = 6.57

d) Solution D : X^- + H⁺ ----> HX

   Initial    0.0025 0.001 0.0015

Change -0.001 -0.001 +0.001

   Equilibrium 0.0015 0.0025

pH = pKa + log [X^-]/[HX] = 6.35 + log (0.0015/0.0025) = 6.35 + log 0.6 = 6.35 -0.22 = 6.13