<p>Using the reaction, he found .01 mol of HF and .006 mol of OH-. Then, you yield .006 mol of F- ions which is a concentration of .15 M with additive volumes.
Next you write the reaction of F- ions with water yielding HF and OH- ions.
Next, find the Kb, because it's a base reaction, which is 1.39 X 10^-11
Set the Kb equal to the concentration of the OH- times the concentration of F-, each of which are X, over the concentration of F-, which is .15. Find the OH concentration take the negative long of that, and then subtract that number from 14 giving you a pH of 8.16.
Not sure if that's right, but that's the way my teacher did it in class today.</p>
<p>BobbyBrown, if you made a mistake in 5di, you lose the point there, but if the same mistake is carried through ii and iii, you shouldn't lose those also</p>
<p>hzbball - the method your teacher used is correct IF you are at the equivalence point (no H+ left). Since there is H+ and F- present, Henderson-Hasselbalch is the way to go.</p>
<p>Im use to 3 decimal places from Calculus, but I think my teacher mentioned if you're off by +-1 decimal, youre fine. Theres like a rule for pH and molarities...like the number of sigfigs in the molarity is the number of decimals in the pH</p>
<p>you have to use SIGNIFICANT FIGURES!!! thats a big part of chem. but, they are really lenient. you can be +/- 1 sig fig, or +/- 2 on pH problems (because its a logarithmic scale).</p>