Question

What volume, in liters, of 1.03 M KOH solution should be added to a 0.110 L solution containing 10.21 g of glutamic acid hydrochloride (Glu, FW= 183.59 g/mol: pKa1= 2.23, pKa2= 4.42, pKa3= 9.95) to get to pH 10.24?.... but whenever i did it the way provided by a different user and got an answer of .036 L it said ....."Incorrect. You did not account for the amount of KOH needed to move glutamic acid through its first two acid dissociations to get to the HGlu– form."

Answer #1

First we calculate the moles of the acid

10.21g/183.59g/mol = 0.055 moles of acid

How many L KOH is that?

0.055moles/1.03 mol/L = 0.054L o 54 mL.

Therefore, it takes 54 mL of the KOH to remove the first H and another 54 mL to remove the second one.

Then the third is the pKa3 where you want the buffer to work.

mols HGlu^- to start = 0.055

mols Glu^2- = 0 at the beginning.

........HGlu^- + OH^- ==> Glu^2- + H2O

I.....0.055.....0.........0

add OH^-..........x...........

C......-x........-x........x

E.....0.055-x....0........x

Then plug into the HH equation

10.24 = 9.95 + log (x/0.055-x)

and solve for x = 0.036 mols KOH added for this part of the
problem

. Convert that to L of 1.03M KOH

0.036mol/1.03mol/L = 0.035 L o 35 mL

and add to L from the first two neutralizations.

54 + 54 + 35 = 143 mL of KOH = 0.143 L of KOH

How many milliliters of 1.72 M KOH should be added to
100. mL of solution containing 10.0 g of histidine hydrochloride
(His·HCl, FM 191.62) to get a pH of 9.30? pKa=9.28

How many milliliters of 1.00 M KOH should be added to 100 mL of
solution containing 10.0 g of histidine hydrochloride [His*HCl =
(HisH+)(Cl-), FM 191.62] to get a pH of 9.30?

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