3. (22 pts) (1) You are titrating acetic acid (0.0200 M)...

3. (22 pts) (1) You are titrating acetic acid (0.0200 M) with NaOH (0.0400 M). What is the pH of the solution after the addition of the following amount of base? $K_a(HAc) = 1.8 \times 10^{-5}$. Show all work. Vol. HAc (mL) Vol. NaOH (mL) pH 20.00 mL 0 mL X = 20.00 mL 8.00 mL Y = 20.00 mL 10.0 mL Z = 20.00 mL 15.00 mL Q = (2) Then roughly construct the titration graph. Label the axes and key points (the deflection points) on the graph.

Transcript

00:01 For this problem we are titrating acetic acid, which is a weak acid, with sodium hydroxide, which is a strong base.

00:10 When they react together they form a salt and water.

00:14 This salt happens to be sodium acetate.

00:18 The sodium comes from the strong base.

00:21 The acetate comes from the weak acid.

00:24 Strong wins, that means that this is a basic salt, which means it acts like a weak base.

00:33 That means we are going to need the kb for when we are at the equivalence point.

00:38 We simply take kw divided by the kb, so 1 times 10 to the negative 14th, divided by 1 .8 times 10 to the negative 5th gives you 5 .6 times 10 to the negative 10th.

00:52 The other pieces of information that you would need would be the equivalence point.

00:56 In this case we have equal concentrations so they will be equal volumes, so 50 milliliters.

01:04 Then the half equivalence point would be half that volume, which would be 25 milliliters.

01:11 We are asked first to find the initial ph of just the acetic acid.

01:18 It is a weak acid, so we will use our ka equals the h plus concentration squared over the concentration of the weak acid.

01:29 Plugging that information in, 1 .8 times 10 to the negative 5th equals the h plus concentration squared over 0 .05.

01:41 Then we will solve this for the h plus.

01:45 1 .8 times 10 to the negative 5th times 0 .05.

01:49 Then we want to take the square root of that and that gives us 9 .49 times 10 to the negative 4th molar.

02:00 It is a concentration and the ph is simply the negative log of that.

02:07 Go ahead and take the negative log of your answer there and you would get 3 .02.

02:17 Next we want to find the ph at the half equivalence point.

02:27 That is the halfway point or the midway point at 25 mils.

02:33 At that point it is lovely because the ph equals the pka, which means it is simply the negative log of your ka value.

02:45 Negative log 1 .8 times 10 to the negative 5th.

02:51 Log of 1 .8 e to the negative 5th equals 4 .75.

03:02 Next we have the ph at the equivalence point, which is a little bit hairier.

03:07 It is not bad, but we need to know how many millimoles of acid and base we have.

03:13 Again we have 50 milliliters of 0 .05 for each...

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