Question

1.800 g of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES–K , MW = 220.29 g/mol) is dissolved in 89.94 mL of water. 29.91 mL of HCl is added to the solution, resulting in a pH of 7.24. Calculate the concentration of the HCl solution. The pKa of ACES is 6.85.

Answer 1

Finding the equlibrium concentrations

ACES^- + H₂O <====> ACES + OH^-

Moles of ACES^- =

$1.80 g ACES^-\times \frac{1 mol}{220.29 g}=8.17 \times 10^{-3} mol$

Volume of ACES^- = 89.94 mL = 0.08994 L

Molarity of ACES- =

$C =\frac{8.17\times 10^{-3}mol}{0.08994L} = 0.0908 M$

pKa = 6.85

pK_b = 14 - 6.85 = 7.15

K_b = 10^-pKb

K_b = 10^-7.15

K_b = 7.08 x 10^-8

$[OH^-]=\sqrt{K_b\times C}$

$[OH^-]=\sqrt{7.08\times 10^{-8}\times 0.0908}$

$[OH^-]=8.02\times 10^{-5} M$

[ACES] = [OH^-] = 8.02 x 10^-5 M

[ACES-] = 0.0908 - 8.02x10^-5 =0.0906 M

Let 'x' be the moles of H⁺ ions added

Total volume of solution after addition of HCl = 89.94 + 29.91 = 119.85 mL = 0.11985 L

Moles of ACES- = Molarity x total volume = 0.0906 x 0.11985 =0.01085 mol

Moles of ACES = 8.02x10^-5 x 0.11985 =9.61 x 10^-6 mol

Using Henderson-Hasselbalch equation

Given pH = 7.24

pOH = 6.76

$pOH = pKb + log\frac{[Conjugate\ acid]}{[Base]}$

$6.76= 7.15 + log\frac{[9.61\times 10^{-6}+x]}{[0.01085-x]}$

$-0.39= log\frac{[9.61\times 10^{-6}+x]}{[0.01085-x]}$

$\frac{[9.61\times 10^{-6}+x]}{[0.01085-x]}= 10^{-0.39}$

$\frac{[9.61\times 10^{-6}+x]}{[0.01085-x]}= 0.407$

x = 0.00313 mol

Moles of HCl = 0.00313 mol

Volume of HCl = 29.91 mL = 0.02991 L

Concentration of HCl =

$=\frac{moles}{volume}= \frac{0.00313 mol}{0.02991L} = 0.105 M$