ph calculation please see details

Chemistry
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Given a 1.00 L solution that is 0.60 M HF and 1.00 M KF, calculate the pH after 0.14 mol NaOH is added, and calculate the pH after 0.22 mol HCl is added to the original solution.

Ka = 7.2×10-4

Mar 9th, 2015

1.  pH after 0.14 mols NaOH added

First, neutralization occurs…

           HF             +      OH     -->            H2O       +        F-

I.....0.60 mols............0.14 mols..................................1.00 mols

C......- 0.14................- 0.14.........................................+ 0.14.......

F.......0.46 mols..........0 mols…………………............1.14 mols.......

We can readily convert mols to molarity since the entire reaction takes place in 1.00L of solvent. Now we determine the amounts of H3O+ present after equilibrium

           HF     +     H2O     <-->         H3O+        +       F-

I.....0.46 M...........................................0 M.............1.14 M

C.......-x................................................+x...................+x.......

E....0.46 - x...........................................x................1.14 + x.......

The equilibrium expression is given by…

7.2E-4 = (x)(1.14 + x) / (0.46 - x)

Using either the quadratic formula or the method of successive approximations, we solve for x…

x = 2.902691E-4 M

Since the hydronium ion concentration is given by x and since pH = -log [H3O+], the pH is given by...

pH =  -log (2.902691E-4)

pH = 3.54 (with sig figs in mind) 


2.  pH after 0.22 mol HCl is added to the original solution

All of the HCl dissociates, giving 0.22 mols of H3O+

The H+ ions are completely neutralized by the F- (basic) ions

           F-            +             H+            -->              HF

I….1.00 mol………..….0.22 mols……………0.60 mols

C…-0.22 mols….........-0.22 mols………….+ 0.22 mols

F…0.78…………………0 mols………………0.88 mols

The equilibrium rxn is…

           HF     +     H2O     <-->         H3O+        +       F-

I.....0.88 M........................................0 M................0.78 M

C.......-x................................................+x...................+x...........

E....0.88 - x.........................................x..................0.78 +x.......

The equilibrium expression is given by…

7.2E-4 = (x)(0.78 + x) / (0.88 - x)

Using the quadratic formula yields…

X = 8.107166E-4 M

Since the hydronium ion concentration is given by x and since pH = -log [H3O+], the pH is given by...

pH =  -log (8.107166E-4)

pH = 3.09 (with sig figs in mind) 


Mar 10th, 2015

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