Question

In: Chemistry

150.0 mL of 0.27 M HF with 230.0 mL of 0.32 M NaF Express your answer...

150.0 mL of 0.27 M HF with 230.0 mL of 0.32 M NaF Express your answer using two decimal places.

Expert Solution

Initial [HF] = 0.27 M

Moles of HF = 0.27*0.150 = 0.0405

Initial [F-] = 0.32 M

Moles of F- = 0.230*0.32 = 0.0736

Total volume = 150 + 230 = 380 mL

Kb of HF = 1.41*10^-11

           F-      +        H2O              HF      +      OH-
Initial      0.0736                 -                  0.0405             0
Change      -x                    -                      +x               +x
Final      0.0736 - x                               0.0405 + x         x

Kb = [HF][OH-] / [F-]

1.41*10^-11 = x*(0.0405 + x) / (0.0736 - x)

Solving this we get:

x = 2.56*10^-11

[OH-] = x / 0.380 = (2.56*10^-11) / 0.380

= 6.736*10^-11 M

pOH = - log [OH-]

= - log (6.736*10^-11)

= 10.17

pH = 14 - pOH

= 14 - 10.17

= 3.83

queen_honey_blossom answered 3 years ago

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