lab 24 - hydrolysis a salt formed between a strong acid and a weak base is an acid salt. ammonia is...
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Lab 24 - HydrolysisLab 24 - Hydrolysis
A salt formed between a strong acid and a weak base is an acid salt. Ammonia is a weak base, and its salt with any strong acid gives a solution with a pH lower than 7. For example, let us consider the reaction
H2O + NH3 NH4+ + OH-
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In the solution, the NH3+ ion reacts with water
(called hydrolysis) according to the equation: NH3
+ + H2O NH4 + OH-.
The acidity constant can be derived from Kw and Kb.
• Ammonia Kb
=1.75e-5
Kw
Ka = ---------------- Kb
= 1.00e-14 / 1.75e-5 = 5.7e-10.
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What is the pH 0f a .1M NaCLO solution if Ka for HCLO 1s 3.0e-8.
step 1 write the net ionic formula• Na + + CLO- + H2O HCLO + Na + OH-
step 2 determine initial, change and equilibrium
[M] of reactants and products
CLO- HCLO OH-
Initial .1 M 0 0
Change xM xM
equilibrium .1M -x XM XM
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What is the pH 0f a .1M NaCLO solution if Ka for HCLO 1s 3.0e-8.
step 3 Find the Kb value
KwKb = ---------------- ------ Ka
1.0e-14 3.3e-7. = ---------------- ------ 3.0e-8.
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What is the pH 0f a .1M NaCLO solution if Ka for HCLO 1s 3.0e-8.
step 3 Find the M of [OH}
[HCLO] [OH-]Kb = ---------------- ------ [CLO]
[X2] 3.3e-7 = ---------------- [0.1]
= 1.8 e -4
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What is the pH 0f a .1M NaCLO solution if Ka for HCLO 1s 3.0e-8.
step 4 Convert [OH] to pOH
-log 1.8 e -4 = 3.74
step 5 Convert pOH to pH14 -3.74 = 10.26
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A. Hydrolysis of saltsA. Hydrolysis of salts
1. Add 5ml of Di water into 6 test tubes
2. Add 3 drops of indicator (as listed on page 270) to each test tube
3. Using the indicators chart on page 262 determine the pH of the water
4. Repeat steps 1-3 using boiled water
5. Repeat steps 1-3 using .1M solutions of the salts listed on page 269
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A. DataA. Data
1. Fill out chart on pg 269 based on the hydrolysis behavior on page 260
2. Determine (H +) for each pH
3. Determine (OH -) for each pH
4. Using data from pg 269fill out chart on page 271
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calculationscalculations
[H+] = 10-pH ( anti log of –pH) [OH-] = Kw / [H+] [OH-] = [1.0 x 10-14] / [H+] Ka or Kb = [M] Products over [M] reactants
• Omit water and assume both products have the same molarity
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Lab 25 – dissociation of a weak acid
Lab 25 – dissociation of a weak acid
General TheoryAccording to the Brønsted-Lowry acid-base theory, the strength of an acid is related to its ability to donate protons. All acid-base reactions are then competitions between bases of various strengths for these protons. For example, the strong acid HCl reacts with water according to Equation [1]:HCl(aq) + H2O(l) H3O
+(aq) + Cl-
(aq)
General
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This is a strong acid and is completely dissociated (in other words, 100 percent dissociated) in dilute aqueous solution. Consequently, the [H3O+] concentration of a 0.1 M HCl solution is 0.1 M. Thus HCl is a stronger acid than water and completely donates a proton to water to form H3O+.
Lab 25
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This is a strong acid and is completely dissociated (in other words, 100 percent dissociated) in dilute aqueous solution. Consequently, the [H3O+] concentration of a 0.1 M HCl solution is 0.1 M. Thus HCl is a stronger acid than water and completely donates a proton to water to form H3O+.
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By contrast, acetic acid, HC2H3O2 (abbreviated HOAc), is a weak acid and is only slightly dissociated, as shown in Equation [2]:
Its acid dissociation constant, as shown by Equation [3], is therefore small• Ka = = 1.8 x 10-5
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titrate the weak acidtitrate the weak acid
If we titrate the weak acid HA with a base, there will be a point in the titration at which the number of equivalents of base is just one-half the number of equivalents of acid present (at ½ Ve).
pH = pKa @ ½ [H+]
By titrating a weak acid with a strong base and recording the pH versus the volume of base added, one can determine the dissociation constant, Ka, of the weak acid
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. C. Determination of pKa of Unknown
. C. Determination of pKa of Unknown
Pipet a 25.00 mL aliquot of your unknown acid solution into a 250-mL beaker and carefully immerse the previously rinsed electrode into this solution.
Measure the pH of this solution. Record the pH in your notebook. Begin your titration by adding 1 mL of your standardized base from a buret and record the volume of titrant and pH.
Repeat with successive additions of 1 mL of base until you approach the end point.• equivalence point--where the graph is steepest• See pg 277
Add 0.1-mL increments of base and record the pH and milliliters of NaOH added untill the pH no longer changes
plot graph to determine ½ equivalence point-
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FormulasFormulas
pH = pKa @ ½ equivalence point-½ equivalence point = 4.3pKa = 4.3Ka = antilog - 4.3 = 5 e-5
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Due Due
Pg 264,265Questions 1-3