0:01

Here at the 25 milliliters of added sodium hydroxide,

Â we have gone beyond the equivalence point.

Â Let's look at a titration curve.

Â At 25 milliliters, we're here at this point.

Â And if you read up, you see that we have passed up the equivalence

Â point by 5 milliliters and the pH shoots up very rapidly.

Â A few extra drops of sodium hydroxide and now we have a strong base

Â in the solution with nothing for it to react with and pH will skyrocket.

Â So, let's go back to our reaction that we have been writing for

Â every point of this titration curve,

Â 0:43

H- plus H2O is my reaction.

Â We will do an ICF table as has been the case in every step,

Â we're not changing the amount of the acid.

Â We've had 4.98 times 10 to the -3 moles every time.

Â For the number of moles of the we'll take the molarity of sodium hydroxide,

Â which is 0.249 moles per liter times the volume.

Â And now it's 0.025 liters and

Â this is going to give me 6.225 times

Â 10 to the minus 3 moles of hydroxide.

Â 1:26

225 times 10 to the minus 3, 0 don't care.

Â Smallest one now is the HA, so

Â we're subtracting 4.98 times 10 to the minus 3 of each of these.

Â And we're producing 4.98 times 10 to the -3 of this.

Â This is going to consume all of the week acid.

Â It is going to leave 1.25 times 10 to the -3 of the hydroxide.

Â And we will have generated 4.98 times 10 to the -3 of the the A- ion.

Â Now at this point,

Â I see a lot of students who want to use the Henderson Hozbog equation

Â because they're used to seeing if there's something on both sides to use that.

Â But the Henderson Hozbog equation is good for a buffer and this not a buffer.

Â It does not have a weak acid in its conjugate base.

Â What does it have?

Â It has a strong base, a strong base and a weak base in solution.

Â Now if you've got a mixture of both a the strong base and a weak base, you can most

Â likely without any trouble just base your problem entirely upon the strong base.

Â So what we need to know is,

Â what is the concentration of After these have reacted and have been added together?

Â We have moles of 1.25 times 10 to the -3 and

Â we have a volume of, if we use this 25

Â milliliters plus the 20 milliliters because it's the total volume in there.

Â So that 45 milliliters or 0.045 liters.

Â This is going to give me the moles of 0.0278.

Â I'm sorry not moles but molarity of the hydroxide.

Â So then we can get the POH and the POH will be pretty high.

Â It's negative log of this value and

Â you take the negative log of that you'll have 1.556 as the POH and

Â in the pH would certainly be 14 minus that 1.556 and

Â that would be a pH of 12.444.

Â So we add the extra hydroxide, the pH rises rapidly,

Â you can base the whole calculation on simply the strong base amount.

Â We didn't even considered this at all.

Â It contributes too few hydroxides to even worry about.

Â