A 605 m solution of potassium hydroxide koh is used to...

A 6.05 M solution of potassium hydroxide (KOH) is used to titrate 205.00 mL of 3.250 M hydrofluoric acid (HF). The Ka for HF is 6.6x10^-4. Assume that volumes are additive. a. What is the pH of the solution before any KOH is added? b. What is the pH after 50 mL of KOH has been added? c. What volume of potassium hydroxide is required to reach the equivalence point? d. What is the pH at the equivalence point?

Transcript

00:01 Okay, so we're reacting some k -o -h with some hcl.

00:08 So let's see what we know about the two of them.

00:11 So we know that this k -o -h has a ph of 13 .29.

00:17 So it's p -o -h.

00:19 We subtract from 14 would be 0 .71.

00:23 And then if we do 10 to the negative.

00:26 Point seven one we'll get the oh h minus concentration which is point one nine five molar okay so there's a concentration of our oh h minus or our k -oh -h for the hcl we've been given the molarity and the volume so we'll go ahead and multiply malarity times liters and that will give us the moles of the hcl or the h plus zero zero six nine one five moles so if we had that many moles of h plus, we must have reacted with that many moles of the oh minus.

01:18 So if we go ahead and divide that by its molarity, which we determined above here, 0 .195 molar, then we should be able to get the volume of the k -o -h that we needed, 0355 liters or 35 .5 milliliters.

01:39 So that would have been the volume of the k -o -h that we needed to add.

01:46 So that was our answer to a.

01:52 So we're trying to find the ph at the equivalence point.

01:56 Well, anytime you add a strong acid and a strong base, there will be no calculations required.

02:04 The ph is always 7 at the equivalence point.

02:09 So then we're going to try to find some concentrations of k -plus and cl -minus...