Titration of strong acid with weak base

reminiscent · Jul 4, 2016

Homework Statement

Homework problem text: 1.665 g of ACES potassium salt (MW = 220.29 g/mol) dissolved in 74.83 mL water, 13.39 mL HCl added, pH 7.42, pKa 6.85; calculate [HCl], with an answer box labeled Number and unit M.

Homework Equations

Henderson-Hasselbalch Equation

The Attempt at a Solution

Handwritten ICE-table working for ACES buffer equilibrium: ACES- + H+ ⇌ ACES, 0.007558 mol, with rows of -x, +x and 0.007558 - x, x, and a log expression beneath.

Sorry for the huge picture, but the number says 0.007558 mol. Is my ICE table correct?
Equation:
7.42 = 6.85 + log((0.007558-x)/x)
(0.007558-x)/x = 3.715
x = 0.00160 mol H+/(0.07483 L + 0.01339 L) = 0.0181 M = [HCl]
It says it is incorrect, though.

Charles Link · Jul 4, 2016

This one is not difficult, but still requires a couple of equations and several algebraic steps. You do need to show the work more clearly for the helper to be able to give good feedback. I think I figured out the solution, but the Physics Forum rules do not allow the helper to simply furnish the solution.

Titration of strong acid with weak base

Homework Statement

Homework Equations

The Attempt at a Solution

Discussion

Similar threads