Determine the pH of a buffer made from 250 ml of 0.50M HF and 150 ml of 0.75 M NaF.
3 answers
What's the problem. Use the HH equation. If you want me to check it let me know what you're using for pKa for HF.
I'm not sure where to start. Is this correct so far?
I first write a balanced chemical equation using HF and water
HF+ H2O <->H3O^+ + F^-
The use the ICE table where the E row is:
HF=0.50-x
H30+= x
F-= 0.75+x
Then the Ka equation
Ka=[H3O+][A-]/[HA] where Ka=(x)(0.75+x)/0.50-x
I first write a balanced chemical equation using HF and water
HF+ H2O <->H3O^+ + F^-
The use the ICE table where the E row is:
HF=0.50-x
H30+= x
F-= 0.75+x
Then the Ka equation
Ka=[H3O+][A-]/[HA] where Ka=(x)(0.75+x)/0.50-x
You are making it far too complicated.
pH = pKa + log (base)/(acid)
millimols base = 150 x 0.75 = ?
millimols acid = 250 x 0.50 = ?
Substitute the millimols into acid/base and solve for pH.
Tehcnically, one can't substitute millimols; i.e., it is supposed to be concentration base and concentration acid BUT note that M (which is the concentration) = millimoles/mL and since the numerator and denominator BOTH have the same mL, the volume cancels and you are left with millimoles. Therefore, millimoles works just fine and that saves the two steps of calculating the concentrations of each.
pH = pKa + log (base)/(acid)
millimols base = 150 x 0.75 = ?
millimols acid = 250 x 0.50 = ?
Substitute the millimols into acid/base and solve for pH.
Tehcnically, one can't substitute millimols; i.e., it is supposed to be concentration base and concentration acid BUT note that M (which is the concentration) = millimoles/mL and since the numerator and denominator BOTH have the same mL, the volume cancels and you are left with millimoles. Therefore, millimoles works just fine and that saves the two steps of calculating the concentrations of each.
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