00:01
For this problem, we're asked to calculate the ph at the equivalence point.
00:04
So at the equivalence point that means that our moles of acid and base are equal to each other.
00:11
And so they will cancel each other out.
00:14
There will be none of either one of those left and all we will have left in solution is this stuff, this nh4cl, which weak base, strong acid, weak base, strong acid, strong wins.
00:26
So this is going to be an acidic salt which will act like a weak acid.
00:34
So we are going to need a ka value.
00:36
So we know kb, ka we can find by taking kw, 1 times 10 to the negative 14th, divided by the kb and that will give us 5 .6 times 10 to the negative 10th.
00:50
And we also need to know the volume for the equivalence point and we find that by multiplying molarity times volume and setting it equal to molarity times volume.
01:01
So 50 times 0 .26.
01:03
I'm going to just write down what that is.
01:05
That's 13 millimoles...