HC?H?O? + H?O 0.150M 13. A 35.0 ml sample of 0.150 M acetic...

HC?H?O? + H?O 0.150M 13. A 35.0 ml sample of 0.150 M acetic acid solution is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added. a. 0 mL (35.ume) (OISUM) (OISUM) (OS.UML) 0.00525 moi HC?H3O? PH=-10g (1.61×10?³) HC?H?O? + OH? ? C?H?O?? + H?O 0.00525 0.002325 1.8x10?? = x² 0.150-x NaOH = (0.0155L)(0.150mol/L) = 0.002325mol acetic acid = (0.035L)(0.150mol/L) = 0.00525mol c. 34.5 mL (0.033L) (0.15GM?) = 0.00525mul 1.8x10?? x² 0.150 x = 1.643x10?³

Transcript

00:01 So we had to calculate the ph of the mixture of 35 mil sample of 0 .15 molar acetic acid with these volumes of 0 .15 molar sodium hydroxide.

00:18 So to start with, the first one, a, 0 mil of the sodium hydroxide.

00:25 So we have 35 mil of 0 .15 molar acetic acid.

00:32 So here we didn't have any sodium nitroxia.

00:35 So we only have the acetic acid.

00:38 And having only the acetic acid acid, we know the concentration already.

00:42 So we can just go ahead and calculate the ph and that'll be negative logarithm of 0 .15 and that's 0 .82.

00:49 Okay.

00:51 And b is we have 17 .5.

00:55 17 .5m .m.

01:00 Of the sodiumide oxide.

01:04 Okay.

01:05 So here it's a little bit different because now we need to know how the reaction happened.

01:10 So we have acetic acid reacting with sodium hydroxide.

01:17 We'll have a formation of the sodium acetate and water.

01:25 Right.

01:25 So the stochymetry for the reaction is one to one.

01:29 Okay.

01:30 So, um, here.

01:34 Here we have the concentration of the sodium my drugs are to be 0 .15 molar.

01:39 So we can determine the number of mole of this.

01:41 So this should be 17 .5 divided by 1000 because we are trying to make this in liters multiply by 0 .15 molar.

01:55 So that will give us a number of mole.

01:56 So this is 17 .5, 1 ,000.

02:09 So this is 0 .0026.

02:15 That is the number of moles of sodium my drugs are here.

02:22 So we have a 0 .002625 more of an a.

02:32 Then we have a 35 .0 mil of 0 .15 molar acetic acid.

02:41 So we're going to do the same thing, 35 25 divided by 1000 multiplied by 0 .15 molar, who give us the number of mol of acetic acid.

02:52 So that's going to be 35 divided by 1000 times 0 .15, and that is 0 .005 .25 more.

03:06 And we can see right from here that we have more number of more of acidic acid than the sodium my drugsyte.

03:14 So that means we're going to have excess of the acetic acid by this among 0 .00525 minus 0 .002625...

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