A) How would you prepare 500. mL of a 0.0333M solution of glutamine with a pH of

Question

A) How would you prepare 500. mL of a 0.0333M solution of glutamine with a pH of 3.73? You have crystalline glutamine hydrochloride, and stock solutions 0.650M HCl and 0.750 M NaOH in the lab to make it.
B) Using activities, calculate the pH of a solution containing 0.0100 M NaOH plus 0.0100 M Mg(No3-)2. Compare you result to what you would get if you ignored activities. A) How would you prepare 500. mL of a 0.0333M solution of glutamine with a pH of 3.73? You have crystalline glutamine hydrochloride, and stock solutions 0.650M HCl and 0.750 M NaOH in the lab to make it.
B) Using activities, calculate the pH of a solution containing 0.0100 M NaOH plus 0.0100 M Mg(No3-)2. Compare you result to what you would get if you ignored activities.
B) Using activities, calculate the pH of a solution containing 0.0100 M NaOH plus 0.0100 M Mg(No3-)2. Compare you result to what you would get if you ignored activities.

Explanation / Answer

A) Prepare 0.0333 M glutamine solution

Glu + H2O <==> Glu- + H3O+

Ka1 = [Glu-][H3O+]/[Glu]

6.76 x 10^-3 = (1.86 x 10^-4)[Glu-]/0.0333

[Glu-] = 1.21 M

moles of Glu- present = moles of NaOH added = 1.21 M x 0.5 L = 0.605 mol

Volume of NaOH to be added = 0.605 mol/0.750 M = 0.81 L

B) activity coefficient using debye-huckel equation

ionic strength (I) = 1/2(0.01 x 2^2 + 2 x 0.01 x 1^2) + 0.01 = 0.04 M

Y[ion] = inv.log[-0.51 x z^2 z sq.rt.(I)/(1+3.3 x r x sq.rt.(I))]

So,

Y[OH-] = inv.log[-0.51 x 1^2 x sq.rt.(0.04)/(1+3.3 x 0.35 x sq.rt.(0.04))]

            = 0.826

pOH = -log[OH-] = -log(0.826 x 0.01) = 2.08

pH = 14 - pOH = 14 - 2.08 = 2.92

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