Question
Asked Nov 13, 2019
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Determine [OH-] and pH of a solution that is 0.140 M in F- .

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Expert Answer

Step 1

Given, concentration of solution with F- is 0.140M.

The reaction of fluoride ion with water can be given as follows:

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F +HO HF+OH° 0 0 Initial 0.140 Change +x+x -х Equilibrium 0.140-x х х

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Step 2

The Kb of HF can be calculated as follows:

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К, оf HF%36.3x10+ к, хк, -к, к, К, к, 1x104 6.3x10 =0.15873x1010.

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Step 3

The expression for Kb of equilibrium and [OH...

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х? К 0.140-x х* 0.15873x1010 0.140-x x-1.50 10M Fromequilibrium [он ]-х [OH1.50x10M pOH-log OH] -log(1.50x10) =5.82

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