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!!

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

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