# Determine the pH of an HF solution of each of the following concentrations?

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Determine the pH of an HF solution of each of the following concentrations?

1. 0.300 M

2. 5.10×10^−2 M

3. 2.60×10^−2 M

In which cases can you not make the simplifying assumption that x is small?

Ka of HF = 3.5*10^-4

• HF is a weak acid so you will have to make an ICE table

HF + H2O —-> H3O+ + F-

0.300

-x……………………+x…..+x

0.300-x……………x……..x

x^2/(0.300-x) = 3.5×10^-4

Assume 0.300-x ~ 0.300

-log[X] to find pH

if x / 0.300 is less than 5%, then your assumption is good. Otherwise, you cannot assume x is small and you will have to solve the quadratic equation.

This will get you going on how to solve the questions.

• Ph Of Hf

• 1) 3.5E-4=x2/(.3-x)

—equation from forming ice table

x=[H+]=.010073

-log.010073 = 1.99684

pH =2

2) pH=2.39

3) pH=2.55

in b and c you cant make that assumption because the actual concentrations produced are more than 5% that the error allows for when using the small number approximation

would you happen to know how to do the buffers in medicine problem??

• pH= -log[concentration]

pH= -log[0.3M]=0.52

pH= -log[5.10 x 10^-2]= 1.29

pH= -log[2.60 x 10^-2]= 1.59

pH=-log[3.5 x 10^-4]= 3.46

These are all acidic because they are under 7. Anything above 7 is basic.

x is smallest in the first one.

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