%3Cp%3ESolve%20an%20equilibrium%20problem%20(using%20an%20ICE%20table)%20to%20ca
ID: 825147 • Letter: #
Question
%3Cp%3ESolve%20an%20equilibrium%20problem%20(using%20an%20ICE%20table)%20to%20calculate%0Athe%20pH%20of%20each%20of%20the%20following%20solutions.%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3EA)%200.15M%20HF%3C%2Fp%3E%0A%3Cp%3EB)%200.15%20NaF%20(This%20one%20I%20got%2C%20it's%208.32)%3C%2Fp%3E%0A%3Cp%3EC)%3Cspan%20class%3D%22c1%22%3Ea%20mixture%20that%20is%200.15%20M%20in%20HF%20and%20.15M%20in%0ANaF%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3E%3Cbr%20%2F%3E%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3EI've%20tried%20looking%20for%20help%2C%20but%20I%20can't%20find%0Aanything%20that's%20right!%20Please%20explain%20how%20to%20do%20it%20too!%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3E%3Cbr%20%2F%3E%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3EThanks!!!!!!%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22replace-insert%20done%20c7%22%20id%3D%0A%22yui_3_11_0_24_1395335129081_62%22%3E%3Cspan%20class%3D%0A%22type-specific%22%3E%3Cspan%20class%3D%22MathJax%20c6%22%20id%3D%0A%22MathJax-Element-8-Frame%22%3E%3Cspan%20class%3D%22math%20c4%22%20id%3D%0A%22MathJax-Span-36%22%3E%3Cspan%20class%3D%22c3%22%3E%3Cspan%20class%3D%0A%22c2%22%3E%3Cbr%20%2F%3E%3C%2Fspan%3E%3C%2Fspan%3E%3C%2Fspan%3E%3C%2Fspan%3E%3C%2Fspan%3E%3C%2Fspan%3E%3C%2Fp%3E%0AExplanation / Answer
A) for a weak acid
HF + H20 ----> H30+ + F-
initial 0.15 , 0, 0
final 0.15-x , x , x
Ka = (F- ) (H30+) / (HF)
6.76 x 10-4 = x 2 / ( 0.15 -x)
x = 0.0097374
pH = -log (H30+)
pH = -log x
pH = -log 0.0097374
pH = 2.011
B )
F- + H20 ----> HF + OH-
Kb = (HF) ( OH-) / (F-)
1.47 x 10-11 = x2 / ( 0.15-x)
x = 1.4849 x 10-6
pOH = -log 0H-
pOH = -log 1.4849 x 10-6
pOH = 5.82
pH = 14 -5.82
pH = 8.18
3) for a acidic buffer
pH = pka + log (salt /acid )
pH = pKa + log ( NaF / HF )
pH = 3.17 + log ( 0.15 /0.15 )
pH =pKa
pH = 3.17
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