Aspirin (acetylsalicylic acid, MW= 180) is a weak acid with pKa of 3.5. If the a

Question

Aspirin (acetylsalicylic acid, MW= 180) is a weak acid with pKa of 3.5. If the aolubility of the compound is 3g/l, what are the concentrations of the protonated and deprotonated forms of a saturated solution at pH 4.5?
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Thank you. Aspirin (acetylsalicylic acid, MW= 180) is a weak acid with pKa of 3.5. If the aolubility of the compound is 3g/l, what are the concentrations of the protonated and deprotonated forms of a saturated solution at pH 4.5?
Show step by step and add explanation if possible.
Thank you. Aspirin (acetylsalicylic acid, MW= 180) is a weak acid with pKa of 3.5. If the aolubility of the compound is 3g/l, what are the concentrations of the protonated and deprotonated forms of a saturated solution at pH 4.5?
Show step by step and add explanation if possible.
Thank you.
Show step by step and add explanation if possible.
Thank you.

Explanation / Answer

Given molar mass of aspirin =180 gm

Solubility of compound is 3 g/ L

So calculate number of moles = weight of compound / molar mass = 3/180 =0.01667 moles

So molarity of given compound =number moles per volume in liter = 0.01667 / 1 = 0.01667 M

Molarity = 0.01667 M (is the concentration of protonated form)

Given Pka =3.5 so convert into Ka

Pka = -log Ka i.e. Ka = 10-PKa so we have

Ka = 10 -3.5 = 3.1622 x 10-4

Dissociation of aspirin is as follows

ASPH            H+ + ASP¯

(Protonated)           ( deprotonated)

Now expression for Ka is

Ka = [H+] [ASP¯] / [ ASPH ]

We know that the aspirin is monoprotic weak acid so [H+] = [ASP¯] so above equation becomes

Ka = [H+]2/ [ ASPH ]

3.1622 x 10-4 =[H+]2/ 0.01667

[H+]2 =5.27 x10-6

[ASP¯] = [H+] =2.29 x 10-3 M (deprotonated form )

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