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

$1.265\ \mathrm g$ of N-(2-acetamido)-2-aminoethanesulfonic acid potassium salt (ACES-K+, $M=220.29\ \mathrm{g/mol}$) is dissolved in $88.42\ \mathrm{mL}$ of water. $27.59\ \mathrm{mL}$ 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 ACES is $6.85$.

I tried setting up the equation as follows: $$\ce{ACES + H3O+ <=> HACES +H2O}$$ The moles of ACES initial is $0.0057\ \mathrm{mol}$ $(1.265/22.290)$ and we are trying to find $x$ the amount of moles of $\ce{HCl}$ $\ce{(H3O+)}$ we have. Therefore, using an ice table $\ce{ACES}=0.00574-x\ \mathrm{mol}$ and $\ce{HCl}= x\ \mathrm{mol}$. 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 $0.0038\ \mathrm{mol}$ of acid divided by total volume = $0.03321\ \mathrm M$. Is that correct?