1a a 200 ml sample of a 0100 m solution of acetic acid...

a. A 20.0 mL sample of a 0.100 M solution of acetic acid, CH3COOH, is titrated with a 0.150 M solution of NaOH. What is the pH after 10.0 mL of NaOH solution have been added? Ka for acetic acid equals 1.8 x 10^–5. a. 3.26 b. 2.87 c. 5.22 d. 7.00 e. 4.27 b. Calculate the pH of the solution at the equivalence point when 25.0 mL of a 0.266 M solution of acetic acid, CH3COOH, is titrated with a 0.336 M solution of sodium hydroxide. Ka for acetic acid equals 1.8 x 10^–5. c. Which of the following indicators is suitable for the titration of 0.10 M NH3 (Kb = 1.8 x 10^–5) with 0.15 M HCl? a. phenolphthalein (Ka = 4.0 x 10^–10) b. bromothymol blue (Ka = 1.0 x 10^–7) c. bromocresol green (Ka = 2.0 x 10^–5) d. thymol blue (Ka = 6.3 x 10^–10) e. cresol red (Ka = 3.5 x 10^–9)

Transcript

0:00 Hello everyone.

00:01 In this question in the first part we have been given a 20 milliliter sample of 20 milliliter sample okay and we have 0 .10 molar concentration of acetic acid that is th3 cooh and it is titrated against 0 .15m of 0 .15 molar concentration of n -a -oh right and we have been given that 10 milliliter volume of noh has been added now we have to find out the ph okay and k a that is the dissociation constant value for this acidic acid is given in the question that is 1 .8 multiply by 10 to the power minus 5 so this is the first part of the question let's see the solution so as we know that the reaction is ch 3 c oh and nuh is added nuh and nuh and it is giving us ch3, coona and water.

01:06 So, equivalence point will be, equivalent point, it will be volume of sample, volume of sample multiply by molarity of sample, molarity of sample divided by molarity of n -a -o -h molarity of base that is enumage now we have to put the values so it is 20 milliliter multiply by 0 .1m divided by 0 .15 molar concentration so it will be equal to 13 .33 milliliter the volume of anoh at the equivalence point will be 13 .33 milliliter after addition of 10 miller of anohage after addition of 10 milliliter of n .a .o .h.

02:08 So, moles of acetate form will be moles of ch3, c -o -o -n -a formed.

02:17 It will be equal to moles of n -o -h added.

02:20 So which means that 10 milliliter multiply by 0 .15 molar concentration.

02:27 So after calculation, it will be equal to 1 .5 multiply with 10 to the power minus 3.

02:33 Moles now we have to calculate the moles of ch3 cooh remained it is equal to initial moles initial moles minus of moles of acetate ion will be formed moles of ch3 coona form so it is equal to 20 milliliter multiply by 0 .1 molar concentration so we can calculate the initial moles and final moles will be the most of acetate ion that is 1 .5 multiply by 10 to the power minus 3.

03:09 So it will be equal to finally 0 .5 multiply by 10 to the power minus 3 moles and by using henderson equation that is ph is equal to pk a plus log moles of salt that is salt concentration divided by moles of acetic acid remained that is the acetic concentration okay so it will be equal to minus log of k a plus log concentration of salt molds of salt will be 1 .5 multiply by 10 to the power minus 3 divided by 0 .5 multiply by 10 to the power minus 3 so as we know that we have been given in the question the k a value that is the dissociation constant that is 1 .5 that is 1 1 .8 multiplied 10 to the power minus 5.

04:07 So after calculation it will be equal to 5 .22.

04:12 So this is the final answer of p .h.

04:14 In the next part of the question also we have to calculate the ph of solution and the concentration at equivalence point has been given.

04:22 So let's see equivalence point will be equivalent point it will be equal to 25 milliliter multiplied by 0 .266 molar concentration divided by 0 .336 molar concentration.

04:43 So it will be equal to 19 .79 milliliter.

04:48 So total volume at equivalence point.

04:52 Total volume at equivalent point.

04:58 It will be equal to initial volume plus of equivalence point.

05:07 Now we have to add these.

05:08 So it is 25 milliliter plus 19 .79.

05:14 25 milliliter plus 19 .79.

05:17 So it will be equal to 44 .79 milliliter.

05:22 So at the equivalence point all the cs3 coh is converted to acetate ion.

05:28 So moles of ch3 coo minus it will be equal to initial moles of cs3 coh.

05:37 Initial moles of ch3 cooh so it is equal to 25 milliliter multiply by 0 .266 molar concentration so after calculation it is equal to 6 .65 multiply by 10 to the power minus 3 moles.

05:58 Let's calculate the concentration of acetate ion that is ch3 coo minus ion so as we know that we have to give the ratio of moles divided by total volume so moles will be 6 .65 multiply by 10 to the power minus t and total volume is 44 .79 so it will be equal to 0 .14847 molar concentration and as we know that ch3 c.

06:28 C .o.

06:29 It is undergoing hydrolysis so in the presence of h2o it is giving us acidicase c .o.

06:36 And oh h minus that is the base so initially 0 .1487 molar concentration of a citate ion will be present so this is 0 and at change of concentration it is minus x this is x and this is also plus x so at equilibrium concentration it will be equal to 0 .1487 minus x and this is plus x and plus kb will be equal to kw divided by ka.

07:10 This is the ratio.

07:11 So kw, as we know that, concentration of ch3, cooh at equilibrium concentration, concentration of oh minus at equilibrium, divided by concentration of acetate ion at equilibrium.

07:30 So it is equal to 10 to the power minus 14 divided by 1 .85 multiple with 10 to the power minus 5...

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