Question

Answer the questions using the data and graph. please show all work. Thank you

Titration of Hydrochloric Acid: 1) Use the buret that is located near the hydrochloric acid container to dispense 25.00 mL of hydrochloric acid solution into a clean dry 100 mL beaker. Record the molarity of this solution on your data sheet. 2) Check to make sure that the volume on your sodium hydroxide buret reads 0.00 mL. If it does not read 0.00 mL, adjust it to do so before you proceed any further. 3) Bring the solution back to your work area. Set up the beaker so that the pH electrode can be immersed and the buret can be over it with its tip slightly below the lip of the beaker. Your instructor may provide you with a magnetic stir bar and hotplate. If so, turn the hotplate on and immediately set the heat temperature to 0 degrees C. This will prevent your titrations from being heated. This is very important a 4) Remove the pH electrode from its storage solution, rinse and blot it and then clamp it so that the electrode is immersed in the acid solution. It's important that the pH electrode does not rest directly on the bottom of the beaker, but that it's immersed in the acid solution. This will allow you to swirl the beaker gently. 5) Click the Red Circle (Preview) button in Capstone. Now double click on the first cell in the first column of the table (under V NAOH). Enter 0.00, and then click the Keep button at the bottom of the Capstone window. The computer will record the pH of your acid solution before you added any base. The second cell in the table under V NaOH will now be highlighted, and a point should appear on your graph. Run #1 Run #1 Run #1 Run #1 Date and T Time (s) V NAOH (mL) pH 15 1.1 5.8 121.2 1.1 10.2 320.6 1.3 15.4 326.8 1.3 16.4 355.4 1.4 17.4 1.4 387.4 18.4 1.5 423.5 19.4 1.5 464 20.4 1.6 520.1 21.4 1.7 570.4 22.4 1.8 612.6 23.4 2 654.6 24.4 2.3 696.7 24.6 C 2.4 723.4 24.8 2.5 795.2 25.1 920.2 25.3 9.1 976.4 25.6 10.5 1014.4 25.8 10.6 1019 26.1 10.9 1051 27.1 11.2 1098.2 28.1 11.2 1144.4 29.1 11.2 1192 30.1 11.2 1242.5 31.1 11.2 1276.2 Second derivative vs Volume of added base 150 100 50 0 25.36 25.31 25.21 25.26 25.16 25.11 25.01 25.06 -50 -100 -150 V (mL) Secind derivative Titration Curves 116 3) What was the pH at the equivalence point? Measured pH Calculated pH Calculations: 4) What was the pH after 30.0 mL of sodium hydroxide solution were added? Calculated pH-- Measured pH-. Calculations: Titration Curves 110 6) Add about 5 mL of base to the acid and allow the mixture to stir for about 30 seconds. Double click in the highlighted cell under V NaOH, read the volume to the nearest 0.1 mL, type it into the field, and then click Keep 7) Add two more 5 mL increments of base, remembering to allow to mix and enter the total volume in the next cell under V NaOH by double clicking on it followed by clicking the Keep button. 8) Add base in 1 mL increments until you reach 24 mL. Don't forget to record each olume in Capstone by double clicking in the appropriate V NaOH cell, typing the volume, and then clicking the Keep button. 9) Now add base in 0.2 mL increments until you reach 26 mL. This helps to define the steepest part of the titration curve better. Again, make sure each volume is recorded in Capstone. 10) After this, add base in 1 mL increments remembering to record each volume in Capstone. Continue to do this until you see that the titration curve that is being plotted levels off and you've got 5 points in the level region of the curve. You should see that the titration curve has gone through its steep portion, which is around the equivalence point. If it hasn't, consult with your instructor. 11) Click the Red Square (Stop) button. Print the plot of the titration curve, and save your data. Also export the data to Excel by clicking on Export under the File menu. Remember that any data you save to the hard drive on the computer will be lost once the laptop has been shut down. You would do well to save your data to a flash drive or to an online storage location. 12) Refill your buret with more base solution back to the 0.00 mL mark in preparation for the next titration. 13) Rinse and blot the pH electrode and move it to its storage solution before disposing the titration mixture in the 100 mL beaker.

Answer 1

12 10 8 6 4 2 0 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 4 -2 -4 -6 Volume of NaOH (in mL) -> <- Hd

In pH vs Volume of base graph, there is sudden change in pH when 25 mL of NaOH has been added to 25 mL of HCl.

So the equivalence point is 25 mL of NaOH.

we can find equivalence point using first derivative graph also.

first derivative graph :

35 30 25 20 15 10 C 15 2b 25 29 10 11 12 13 14 16 17 18 19 20 21 24 26 27 28 30 volume of NaOH op un ApH/AV

So, equivalence point = 25.2 mL

1) to calculate pH of HCl before addition of base

Since HCl is a strong acid, it will dissociate completely to give H⁺ and Cl^- ions

HCl $\rightarrow$ H⁺ + Cl^-

So, [HCl] = [H⁺]

we have 0.1 M HCl

so, [H⁺] = [HCl] = 0.1

pH = - log [H⁺] = -log (0.1) = - (-1)

pH = 1

Calculated pH = 1 Measured pH = 1.1

2) To calculate pH on addition of 10 mL of 0.1 M NaOH in 25 mL of 0.1 M HCl

Volume of HCl solution = 25 mL = 25/1000 = 0.025 L

Moles of HCl = Volume of HCl solution x Molarity of HCl

= 0.025 x .0.1 = 0.0025 mol

Volume of NaOH solution = 10 mL = 10/1000 = 0.01 L

Moles of NaOH = Volume of NaOH solution x Molarity of NaOH

= 0.01 x 0.1 = 0.001 mol

NaOH and HCl will react with each other and neutarlization will take place as per the following reaction:

HCl + NaOH $\rightarrow$ NaCl + H₂O

according to the reaction, 1 mol of NaoH reacts with 1 mol of HCl

so, 0.001 mol of NaOH will react with 0.001 mol of HCl and neutralize it

Amount of HCl left unneutralized = total moles of HCl - moles of HCl neutralized by NaOH

= 0.0025 - 0.001

= 0.0015 mol

So, the pH of the solution will be due to the unneutralised HCl

Moles of unneutralized HCl = 0.0015 mol

Total volume of the solution = 25 mL + 10 mL = 35 mL

= 35/1000 = 0.035 L

Molarity of unneutralised HCl = moles / volume of solution = 0.0015 / 0.035

= 0.04286 M

[H⁺] = [HCl] = 0.04286 M

pH = -log [H⁺] = -log (0.04286)

= -(-1.3679)

= 1.3679 = 1.37

Calculated pH = 1.37

Measured pH = 1.3

3) to calculate the pH of the solution at equivalence point

At equivalence point, we have added 25 mL of 0.1 M NaOH is added to 25 mL of 0.1 M HCl

Volume of HCl solution = 25 mL = 25/1000 = 0.025 L

Moles of HCl = Volume of HCl solution x Molarity of HCl

= 0.025 x .0.1 = 0.0025 mol

Volume of NaOH solution = 25 mL = 25/1000 = 0.025 L

Moles of NaOH = Volume of NaOH solution x Molarity of NaOH

= 0.025 x 0.1 = 0.0025 mol

NaOH and HCl will react with each other and neutarlization will take place as per the following reaction:

HCl + NaOH $\rightarrow$ NaCl + H₂O

according to the reaction, 1 mol of NaoH reacts with 1 mol of HCl

so, 0.0025 mol of HCl will react with 0.0025 mol of NaOH and neutralize it

Amount of NaOH left unneutralized = total moles of NaOH - moles of NaOH neutralized by HCl

= 0.0025 - 0.0025

= 0 mol

So, at equivalence point, there is complete neutralization of HCl by NaOH and only NaCl and H₂O are present in the solution

so, pH will be due to water and hence ionization of water will become significant , and at 25^oC, [H⁺] = [OH^-] = 10^-7

pH = -log [H⁺]

= -log(10^-7)

= -(-7)

Calculated pH = 7

4) To calculate pH of the solution on addition of 30 mL of 0.1 M NaOH in 25 mL of 0.1 M HCl

Volume of HCl solution = 25 mL = 25/1000 = 0.025 L

Moles of HCl = Volume of HCl solution x Molarity of HCl

= 0.025 x .0.1 = 0.0025 mol

Volume of NaOH solution = 30 mL = 30/1000 = 0.03 L

Moles of NaOH = Volume of NaOH solution x Molarity of NaOH

= 0.03 x 0.1 = 0.003 mol

NaOH and HCl will react with each other and neutarlization will take place as per the following reaction:

HCl + NaOH $\rightarrow$ NaCl + H₂O

according to the reaction, 1 mol of NaoH reacts with 1 mol of HCl

so, 0.0025 mol of HCl will react with 0.0025 mol of NaOH and neutralize it

Amount of NaOH left unneutralized = total moles of NaOH - moles of NaOH neutralized by HCl

= 0.003 - 0.0025

= 0.0005 mol

So, the pH of the solution will be due to the unneutralised NaOH

Moles of unneutralized NaOH = 0.0005 mol

Total volume of the solution = 25 mL + 30 mL = 55 mL

= 55/1000 = 0.055 L

Molarity of unneutralised NaOH = moles / volume of solution = 0.0005 / 0.055

= 0.009 M

[OH^-] = [NaOH] = 0.009 M

pOH = -log [OH^-] = -log (0.009)

= -(-2.04))

= 2.04

pH + pOH = 14

pH = 14 - 2.04 = 11.96

Calculated pH = 11.96

Measured pH = 11.2