anonymous
  • anonymous
The compound para–aminobenzoic acid (PABA) is a powerful sun–screening agent whose salts were once used widely in sun tanning and screening lotions. The parent acid, which we may symbolize as H–Paba, is a weak acid with a pKa of 4.92 (at 25 °C). What are the [H+] and pH of a 0.030 M solution of this acid? Please help!!
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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aaronq
  • aaronq
So write an equation for the dissociation of the acid, then use it to write an equilibrium expression. \(\sf pK_A=-log(K_A)\)
anonymous
  • anonymous
i thought that you would find H+ by multiplying 4.92 and 0.030 and then for the ph you would solve for x?
aaronq
  • aaronq
it's not too far off. multiply the Ka by 0.03 then take the squared root, then you have [H+]. Next take the negative logarithm of [H^+] then you have the pH

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anonymous
  • anonymous
i got -.021 for H+? is that right?
aaronq
  • aaronq
no it can't be negative
anonymous
  • anonymous
okay so would it be .021 then?
anonymous
  • anonymous
i did the -log of 4.92 and then i multiplied it by the0.030
aaronq
  • aaronq
the first step is wrong, the formula is \(\sf pK_A=-log(K_A)\), so then \(\sf K_A=10^{-pK_A}\)
anonymous
  • anonymous
okay i got 3.61?
aaronq
  • aaronq
what is that?
anonymous
  • anonymous
for H?
aaronq
  • aaronq
Does that make sense to you? the initial concentration of the acid is 0.030 M
anonymous
  • anonymous
yes
aaronq
  • aaronq
no it can't be higher than the concentration of the acid
aaronq
  • aaronq
did you do what i said previously? "multiply the Ka by 0.03 then take the squared root, then you have [H+]"
anonymous
  • anonymous
yeah you said ka was equal to 10^-4.92? and then i multiplied it by 0.030? and then i took the square root of it and got 6.00x10^-4
aaronq
  • aaronq
yeah, that answer is right
aaronq
  • aaronq
it's not what you wrote before though
anonymous
  • anonymous
yeah i forgot to take the square root
anonymous
  • anonymous
and then for the ph i get 3.22?
aaronq
  • aaronq
yep that's right
anonymous
  • anonymous
yay! thanks!!
aaronq
  • aaronq
no problem !

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