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 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.
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