Find the concentration of the HCl solution when mixed with 1.265g of ACES-K+

Question

$\pu{1.265 g}$ of N-(2-acetamido)-2-aminoethanesulfonic acid potassium salt ($\ce{ACES^-K+}$, $M=\pu{220.29 g/mol}$) is dissolved in $\pu{88.42 mL}$ of water. $\pu{27.59 mL}$ of $\ce{HCl}$ is added to the solution, resulting in $\mathrm{pH}$ of $6.54$. Calculate the concentration of the $\ce{HCl}$ solution. The $\mathrm pK_\mathrm a$ of $\ce{ACES}$ is $6.85$.

I tried setting up the equation as follows: $$\ce{ACES + H3O+ <=> HACES +H2O}$$

The moles of ACES initial is $\pu{0.0057 mol}$ $(1.265/22.290)$ and we are trying to find $x$ the amount of moles of $\ce{HCl}$ $\ce{(H3O+)}$ we have. Therefore, using an ice table $\ce{ACES}=0.00574-x~\pu{mol}$ and $\ce{HCl}= x~\pu{mol}$. When you plus into the equation $$\mathrm{pH}=\mathrm pK_\mathrm a+\log(\text{base}/\text{acid})$$ I did $6.54=6.85+\log(0.00574-x/x)$ and got $\pu{0.0038 mol}$ of acid divided by total volume = $0.03321\ \pu M$. Is that correct?

Answer 1

Let us call the potassium salt "A-" and its corresponding acid for "HA". The total volume of the final solution is 88.42 + 27.59 = 116.01 ml. The final concentration of [HA] + [A-] = 1.265/(220.29 *0.11601) = 0.049499 (M). [K+] = [HA] + [A-] = 0.049499 (M).

The equilibrium equation gives: 6.54 = 6.85 + log ([A-]/[HA]) => [A-]/[HA] = 0.48977 => [A-] = 0.48977 * [HA]. => [HA] + 0.48977 * [HA] = 0.049499 => [HA] = 0.03322 (M) and [A-] = 0.016279 (M).

The charge balance is: [H3O+] + [K+] = [OH-] + [Cl-] + [A-] At pH = 6,54 [H3O+] = 10^-6.54 and [OH-] can be neglected in comparison with [A-]. => [Cl-] = [H3O+] + [K+] - [A-] = 0.03322 (M).

The amount of chloride ions in the final solution will be 0.03322 * 0.11601 = 0.003853 mole. Thus, the concentration of the hydrochloric acid used will be: (0.003853 * 1000) /27.59 = 0.1396 or about 0.140 M.