Consider the titration of 1000 ml of 0016 m hocl with 00400...

consider the titration of 100.0 mL of 0.016 M HOCl with 0.0400 M NaOH how many milliliters of 0.0400 M NaOH are required to reach the equivalence point? calculate the pH after the addition of 10.0 mL of 0.0400 M NaOH calculate the pH halfway to the equivalence point calculate the pH at the equivalence point

Transcript

00:01 Question 15 .16 is a rather lengthy problem.

00:05 It has three parts, three different titration calculations.

00:09 In addition to also calculating the equivalence point volume, which is necessary in order to make sense out of the three additional titration volumes they give you.

00:20 To solve for the equivalence point volume, you will take the initial volume of the analyte that you have, multiplied by its concentration, to figure out the moles of the amount of the amount, the h .o .c .l.

00:33 That you have, then recognize that the reaction of h .o .c .l.

00:38 With n .o .h.

00:39 Is a one -to -one molar reaction.

00:41 You will then have at this point the moles of strong base that you need to add.

00:47 Using the molarity of the strong base, you can calculate the moles of strong base needed.

00:53 And then, i'm sorry, using the molarity, you can calculate the leaders of strong base needed, then you can multiply by a thousand in order to get the volume in milliliters of strong base needed.

01:07 So the equivalence point volume is 40 .00 milliliters.

01:13 So after the addition of 10 milliliters, we know that we are pre -equivalents, because 40 is the equivalents.

01:21 When we are pre -equivalents, with a weak acid being titrated with a strong base, we have a buffer solution.

01:28 So we need to the moles of weak acid left in solution and the moles of weak base that we have formed.

01:36 The moles of weak acid left in solution is going to be equal to the moles of weak acid we started with, molarity times volume, multiplied by the moles of weak acid that reacted.

01:47 The moles of weak acid that reacted will be equal to the moles of strong base formed, so molarity times volume.

01:54 Then the moles of weak base formed will be equal to the moles of strong base added.

01:58 So the moles of strong base added again is 10 milliliters multiplied by the concentration of that strong base .04.

02:06 Now knowing the moles of weak acid and its conjugate base left in solution, we can use the henderson -hasselmaltz equation.

02:15 Ph is equal to p -ka, the negative log of the k -a value that was given to us, plus the log of the moles of the base over the moles of the acid, which we just calculated up here, we'll then calculate that ph of 6 .98...

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