V.Preparing a Buffer Solution 19. Weigh a clean 150-mL beaker to the nearest 0.1
ID: 1049392 • Letter: V
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V.Preparing a Buffer Solution 19. Weigh a clean 150-mL beaker to the nearest 0.1 g NOTE: Ask your laboratory instructor to verity the mass of NaC2H302 you should in Step 20 20. Add 2.0 g of solid anhydrous NaC2H,O2 to the beaker. CAUTION 6.0MHC2H202 is corrosive. Prevent eye, skin, and clothing contact. If you spill any acid, immediately notify your laboratory instructor. 21. Remove the beaker from the balance. Using a 10-mL graduated cylinder, add 4.0 mL of 6M HC2H302 to the beaker. 22. Using a 50-mL graduated cylinder, add 46 mL distilled water to the beaker. Stir the solution until the NaC2H3O2 has completely dissolved This solution will contain 2.4x 10 mol each of NaC2H,02 and HC2H,O2 in 50 mL of solution. VI.Observing pH Changes in 23. Label four clean, dry 50-mL beakers ",and "4". Water and Buffer Solutions 24. Using a 50-mL graduated cylinder, pour 25 mL of distilled water into upon Addition of HCland beakers 1 and 3. NaOH Solutions 25. Using a graduated cylinder, add 25 mL of the buffer solution each to beakers 2 and 4.Explanation / Answer
Theoretical [H3O^+] :
Distil Water : 1.62*10^-7
organic buffer solution :1.99*10^-5
Distill water with added HCl = 0.074M
Buffer with HCl=3.8*10^-5 M
Distill water with NaOH = 1.55*10^-13 M
Buffer with NaOH =1.15*10^-5 M
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Calculated pH :
(1) Distil water = 7
(2) pH of organic buffer solution can be calculated by using Hinderson hasselbalch equation.
Molarity of acetate salt and acetic acid = 2.4*10^-2moles *1000mL/50mL =0.48 M
pH = pKa + log[acetate/acetic acid] = 4.74 + log[0.48/0.48] = 4.74
(3)pH of distill water after added HCl :
moles of HCl in 0.5 mL = 6M *0.5mL/1000mL = 0.003 moles
Molarity of HCl solution = 0.003 *1000mL/25mL = 0.012 M
pH = -log[0.012] = 1.92
[H3O+] = 10^-pH = 0.012 M
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4) Buffer solution with added HCl.
Moles of acetate in 25mL = 1.2*10^-2 M
Moles of acetic acid = 1.2*10^-2 M
Moles of acetate after adding HCl = 1.2*10^-2-0.003 = 0.009 moles
Molarity = 0.009*1000mL/25.5 mL = 0.353 M
Moles of acetic acid after adding HCl =1.2*10^-2 +0.003 = 0.015 moles
Molarity of acetic acid after adding HCl = 0.015*1000mL/25.5mL = 0.59 M
pH = pKa + log [acetate/acetic acid ]
= 4.74 + log[0.353 M/0.59 M] = 4.52
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(5) pH after adding NaOH to Distil water
pH of distill water after added HCl :
moles of NaOH in 0.5 mL = 6M *0.5mL/1000mL = 0.003 moles
Molarity of NaOH solution = 0.003 *1000mL/25.5mL = 0.012 M
pOH = -log[0.012] = 1.92
pH = 14-1.92 = 12.08
[H3O+] = 10^-pH = 8.32*10^-13 M
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Buffer solution with added NaOH.
Moles of acetate in 25mL = 1.2*10^-2 M
Moles of acetic acid = 1.2*10^-2 M
Moles of acetate after adding NaOH = 1.2*10^-2+0.003 = 0.015 moles
Molarity = 0.015*1000mL/25.5 mL = 0.59 M
Moles of acetic acid after adding HCl =1.2*10^-2 -0.003 = 0.009 moles
Molarity of acetic acid after adding HCl = 0.009*1000mL/25.5mL = 0.353 M
pH = pKa + log [acetate/acetic acid ]
= 4.74 + log[0.59 M/0.353] = 4.96
[H3O+] = 1.096*10^-5 M
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