Question

In: Chemistry

1.152 g of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES–K , MW = 220.29 g/mol) is dissolved in...

1.152 g of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES–K , MW = 220.29 g/mol) is dissolved in 87.80 mL of water. 10.56 mL of HCl is added to the solution, resulting in a pH of 7.45. Calculate the concentration of the HCl solution. The pKa of ACES is 6.85.

Expert Solution

m = 1.152 g ACES

MW = 220.29

V = 87.80 ml

V2 = 10.56 HCl

ph =7.45

calculate HCl in solution

pKa 6.85

Ka = 10^-6.85 = 1.41*10^-7

Kb = 7.09*10^-8

if PH = 7.45 then pOH = 6.55

this means [OH-] = 10^-6.55 = 2.81*10^-7

From here

Kb = [B+][OH-]/[B]

x = [OH-]= [B+] = 2.81*10^-7

[B] = M - x =

solve for [B]

Kb = [B+][OH-]/[B]

7.09*10^-8 = (2.81*10^-7)*(2.81*10^-7) / (M -2.81*10^-7)

Solve for M

(7.09*10^-8) * (M -2.81*10^-7) = 7.89*10^-14

(M -2.81*10^-7) = 1.11*10^-6

M = 1.39*10^-6

That is, moles per volume, we hav ea volume of Vt = 87.80 + 10.56 = 98.36 V

find moles

n = M*V = 1.39*10^-6*98.36 = 1.367*10^-4 mmol

This is the amout AFTER the addition of HCl

find it before the addition

n1 = m/MW = 1.152 / 220.3 = 0.005229

Change = 0.005229 - 1.367*10^-7 = 0.005229 moles of change

since 1 mol of ACES reacts with 1 mol of acid then that was the amount of moles in HCl solution

M acid = mol acid/ V acid

M = 0.00509 / 10.56 = 0.48 M

M = 0.48 M HCl

queen_honey_blossom answered 4 months ago

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