Asked by Anonymous
A bottle of vinegar is found to have a pH of 2.90. Calculate the molar concentration of acetic acid in this vinegar bottle (assume that acetic acid is the only acid present in this aqueous solution). The pKaof acetic acid is 4.76.
Answers
Answered by
bobpursley
so you know the pK and the pH, looking for Molarity M.
The amount of hydrogen ion (x) is the original Molarity M minus the amount dissociate.
Ka=x^2/(M-x)
or Ka(HA-x)=x^2 or HA-x=x^2/Ka
now, lets swithc to logs in the x term..
HA-10^<sup>-pH</sup> = (10^<sup>-pH</sup>)^<sup>2<</sup> /K
now with some algebra, solve for HA. You know pK=4.76, so you then know K= 10^(-4.76)== 1.73780083e-5
HA= (10^<sup>-pH</sup>)^<sup>2<</sup> /K + 10^<sup>-2pH</sup>
That is the equalibirum cocentration
original cocentration M then Add 10^{-ph} for the concentration of the acid before dissociation.
The amount of hydrogen ion (x) is the original Molarity M minus the amount dissociate.
Ka=x^2/(M-x)
or Ka(HA-x)=x^2 or HA-x=x^2/Ka
now, lets swithc to logs in the x term..
HA-10^<sup>-pH</sup> = (10^<sup>-pH</sup>)^<sup>2<</sup> /K
now with some algebra, solve for HA. You know pK=4.76, so you then know K= 10^(-4.76)== 1.73780083e-5
HA= (10^<sup>-pH</sup>)^<sup>2<</sup> /K + 10^<sup>-2pH</sup>
That is the equalibirum cocentration
original cocentration M then Add 10^{-ph} for the concentration of the acid before dissociation.
Answered by
bobpursley
check my math, it is a pain to type in HTML to get superscripts, then proof it.
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