The end point or equivalence point of your acid base...

The end point (or equivalence point) of your acid / base titration between HCl and NaOH occurred when: the acid and the base neutralized each other. the phenolphthalein indicator turned a faint pink color. the moles of H+ = the moles of OH-. Both A and B are correct. A, B, and C are correct. For number 18-20 Consider the titration of 50.0 mL of 0.35 M NaOH with 0.200 M HCl. Question 18 Calculate the pH at initial stage. pH =1.34 pH = 7.00 pH = 11.12 pH = 13.54 Question 19 Calculate the pH after addition of 50 mL HCl. pH = 1.12 pH = 12.88 pH = 13.00 pH =13.88 Question 20 Calculate the pH after addition of 100 mL HCl. pH = 1.78 pH = 2.34 pH = 3.15 pH = 12.22

Transcript

00:01 But for this question, they don't make it very clear whether we're starting with the acid and adding the base or we're starting with the base and adding the acid.

00:08 They simply state that we're doing an acid slash base titration with hcl and sodium hydroxide.

00:15 So i'm going to make the assumption we're starting with the acid and we're adding the base.

00:20 If that's the case, then phenothelene will turn pink once we reach the equivalence point as opposed from going from, as opposed changing from pink to clear.

00:31 Also at the equivalence point, we will have neutralized the acid and base.

00:35 They would have neutralized each other.

00:37 And the amount of h plus, the moles h plus would equal the moles of hydroxide.

00:41 So all three of them are correct, a, b, and c.

00:45 To calculate the ph at the beginning of the titration, before we add any hcl, see, so now they've flipped it, we're starting with base and adding hcl.

00:56 In this case, it's going to go from pink to colorless.

01:00 But anyway, we need to know the hydronium concentration.

01:05 Ph is always the negative log of the hydrogenium concentration.

01:09 The hydrogenum concentration is simply the hydroxide, coming from sodium hydroxide, 0 .35, divided into kw.

01:17 This gives us hydronium.

01:19 We take the negative log, we get a ph of 13 .54.

01:23 Then to calculate the ph after the addition of 50 milliliters of 0 .2 molar hcl, well, we started with 50 milliliters of 0 .35 molar sodium hydroxide, so we would have not added enough hcl to neutralize all of the base if we add 50 milliliters of only 0 .2 molar.

01:43 So my point is there is excess hydroxide.

01:46 So to calculate ph, we again need to take the negative log of the hydrogenium concentration, but we first need to figure out what the excess molarity of hydroxide is.

01:58 To calculate that, we will take the volume of sodium hydroxide 50 milliliters, expressed as liters as 0 .05, multiplied by its molarity.

02:09 This will give us moles hydroxide that we start with...

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