<p>Thirty mL of 0.100M HC2H3O2, is tritrated with NaOH. If the total volume of the solution at the eequivalence ppoint is 45 mL what is the pH of the solution?</p>
<p>that means theres 15 ml added of NaOH to reach the equivalence point</p>
<p>Since at the equivalence point
Since this is at the equivalence point it should be the pH of a salt, since it is weak-strong titration (determined by the acetate ion concentration)</p>
<p>so to calculate pH of a salt, which is the equation</p>
<p>C2H3O2- + H2O equilib HC2H3O2 + OH-</p>
<p>the equilibrium expression becomes</p>
<p>Kb = [HC2H3O2][OH-]/[C2H3O2-]</p>
<p>Kb = Kw/Ka
Kb = 10^-14/1.8x10^-5 = x^2/[C2H3O2-]</p>
<p>Solve for moles of HC2H3O2: M x V </p>
<p>.1 M x .03 = .003 moles C2H3O2- at equivalence point</p>
<p>.003/.045 = new molarity = .06667</p>
<p>pH = 14+log(sqrt(.06667 x (10^-14/1.8x10^-5))) (algebraic manipulations here)</p>
<p>pH = 8.7843</p>
<p>"Kb = Kw/Ka
Kb = 10^-14/1.8x10^-5 = x^2/[C2H3O2-]</p>
<p>Solve for moles of HC2H3O2: M x V "</p>
<p>This is where I get lost. What is the concentration of C2H302- which allowed you to solve for X?</p>
<p>The concentration of acetate is .003 moles (from the original concentration - at the equivalence point all of the acetic acid is converted into acetate) divided by the milliliters of solution. That is, .003/.043 = .06666666 bar.</p>
<p>Oh nvm. I understand now. I was treating this as a problem harder than it actually was (titration types).</p>
<p>God, now I don't understand it again. By equivalence you mean equilibrium?</p>
<p>"Solve for moles of HC2H3O2: M x V</p>
<p>.1 M x .03 = .003 moles C2H3O2- at equivalence point</p>
<p>.003/.045 = new molarity = .06667"</p>
<p>By titration, does that mean all of the original acid is gone? Acetic acid barely disassociates (relatively), so not all of the C2H3)2- formed after reacting with NaOH would stay that way. A lot would react back to acetic acid.</p>
<p>First of all I just want to say that lollerpants method and answer are absolutely correct. Except for sig figs, but nobody cares about those.</p>
<p>Anyway, afruff, yes, a titration to the equivalence point means that all the original acid has reacted:</p>
<p>HC2H3O2 + OH- --> C2H3O2- + H2O</p>
<p>What you are saying (that part of it would return to being acetic acid) is exactly what the next part of the problem entails. Remember, we are finding the Kb of the acetate ion, which is an equilibrium for the reaction:</p>
<p>C2H3O2- + H2O <---> HC2H3O2 + OH-</p>
<p>In other words, the Kb takes into account how much of it returns to acetic acid.</p>
<p>I just wanted to say that the reason I didn't understand this type of problem is because we haven't gone over it yet in our class.</p>
<p>My teacher said that there would be no Strong-Weak titration questions on the AP test, just so everyone knows.</p>
<p>There have definitely been Strong-Weak titration free response questions in the past. What makes your teacher say there won't be?</p>