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I have my acids and bases test tomorrow, and i need help with a few types of problems

  • Thread starter pakmingki
  • Start date
  • #1
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im looking at some review problems my teacher gave me today, and i can pretty much do most of them. however, there is this one i really have no idea what to do. Its a 2 part problem.

1a.What is the pH of a 1000 ml of a 0.1 M acetic acid solution, given pKa = 4.8?
1b. what is the pH of 1000ml of 0.1 M acetic acid solution after you have added 5 ml of 10.0 M NaOH, assuming net volume change is negligible

so, for the first part, i think i got it wrong. But here is what i did.
First i found the Ka with
pKa = 4.8 = -log[Ka]
10^-4.8 = Ka
Ka = 1.58 x 10^5

so then i wrote out the equilibrium reaction.
C2H4O2 + H20 > C2H3O2 + H30
where everything is aqueous except H20, and the products on the right side are - and + ions, respectively.
so then
Ka = [x][x]/0.1 - x, where is x is the concentration of H30 and C2H3O2.
solving for x gives 1.25 * 10^-3

so, [Hplus] = 1.25 * 10^-3 M
then, i did this step
(1.25 x 10^-3 Hplus ions/L) * 1000ml * 1L/1000ml gives 1.25 x 10^-3 Hplus ions

pH = -log[Hplus] = 2.903

i really need help, im pretty sure i made some erroneous steps.

and for part b, i really didnt have any idea how to start. I just started by finding the number of mols of OHminus, but i dont think that does anything.
So, can someone help me do both parts?
thanks
 

Answers and Replies

  • #2
symbolipoint
Homework Helper
Education Advisor
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Hi, Pakmingki,
You essentially have the right idea. In part A, you obtain about the same result whether a more full quadratic form is used or the simplification. In part B, the concentration of acid is essentially cut in half.

This is a BASIC program which can give the results that you want; if you want a more full explanation, write a private message:
let Ka=1.58*10^(-5)
let M=0.1
'note, let M=0.05 if want to
'cut concentration in half, as if
'added 5ml of 10M NaOH

let x=Ka+sqr(Ka*Ka+4*M*Ka)
let x=x/2

pH=-1*log(x)/log(10)

print "concentration ";x
print "pH ";pH

END
You get a pH of about 3 for the addition of caustic soda, a pH of a bit less than 3 for just the acid before the addition of caustic soda.
 
  • #3
symbolipoint
Homework Helper
Education Advisor
Gold Member
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Excuse me, but what I wrote for part B may be inaccurate. Better, to take this form instead for the addition of NaOH:

if x= molarity of hydronium,
Fs = formal salt concentration
Fa = formal acid concentration

x(Fs + x)/(Fa - x) = Ka

Now, just find x; this is a quadratic equation, so no real trouble. My little BASIC program is fine for little or no salt, but is not necessarily so good when a large portion of salt (neutralized acid) is present.
 
  • #4
Borek
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