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

$$\pu{1.265 g}$$ 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?