Buffers Question ,need help plz

[integralx2]

I am having an issue understanding. For instance if I have a 500ml solution of 0.1M formate buffer (HCOO-) at pH =3.75. Then i add to it 5ml 1.0M NaOH. How do i get the pH after adding the strong base.

I am trying to do this dont know if right. But i have 500ml solution of 0.1m , that is .25 moles of HC00- . Then I am told my pH is 3.75. Knowing the pH=-log[h+] i get .177828x10^3 M as my concentration of H+. So i have a 500ml x .177828x10^3 M = .0899 moles of H+. Now I am stuck. The equilibrium equation would look like (if i am correct)

HCOO- + H20 ---> HCOOH + OH-

That H+ concentration is that the same thing as my initial HCOOH concentration ? or is that the H20 concentration since it would act like the acid ? Also when i ADD NaOH , it dissociates completely and i have OH- . Do i just add then HCOO- M + OH- M = total HCOO- M ?

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[Pons Asinorum]

I am having an issue understanding. For instance if I have a 500ml solution of 0.1M formate buffer (HCOO-) at pH =3.75. Then i add to it 5ml 1.0M NaOH. How do i get the pH after adding the strong base.

It's a BUFFERED solution. When you add a small amount (5 ml/500 ml) of an acid or a base, then the pH change will be very small (assuming that it is buffered near it's half-equivalence point, pKa = pH, all the prep books have pretty good explanations of this). Adding a small amount of strong base to a large amount of buffered solution will cause a very small increase in pH. You don't need to do math; the answer should be obvious from inspection of the answer choices.

 

[DougTPR]

You don't need to do math; the answer should be obvious from inspection of the answer choices.

Sometimes this is what I say to my students, question after question. Always remember that you don't have to be RIGHT, you only have to be close so estimates are always the easier way to go.