PHCM223 Midterm Revision SS16 443

Pharmaceutical Analytical Chemistry (PHCM223-SS16)

Midterm

REVISION LECTURE

Dr. Dr. Rasha Ras ha Hanafi PHCM223,SS16, 26-03-2016

Midterm revision, Dr Rasha Hanafi.

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LEARNING OUTCOMES O UTCOMES By the end of this session, the student should: •

Comprehend pH relationships.

•

Solve pH problems for acids, bases, salts, sal ts, buffers. buffers.

•

Select the right indicator for an acid base titration

•

Select the suitable buffer buffer for an application

•

Calculate the concentrations of the components of a buffer

PHCM223,SS16, 26-03-2016


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pH RULES pH Strong acids = - log[H+] pOH Strong bases = =log [OH-] Kw = [H+][OH-]= 1.0x10-14 pH + pOH=14 

K a [ weak acid ]

pH of weak acid = -log

at 25°C

k a

Calculation of the pH of a weak base: 1. Find the [OH-] from the pKb value 2. From the [OH-] find the pOH 3. Find the pH from: pOH + pH = 14





[ H ][CH 3COO ] [CH 3COOH ] 

k b





[ NH 4 ][OH ] [ NH 3 ]

For weak polyprotic acids, Ka1>Ka2>Ka3 Salts made of cations of strong bases and the anions of strong acids have no effect on [H +] when dissolved in water.

SALTS OF WEAK ACIDS IN WATER GIVE BASIC SOLN. SALTS OF WEAK BASES IN WATER GIVE ACIDIC SOLN. Buffers pH = pK a + log PHCM223,SS16, 26-03-2016

[base ] [acid]

When [Base] = [acid]  pH = pKa pH= pKin + log


[]

−

[ ]

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TITRATION CURVE STRONG ACID – STRONG BASE

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TITRATION CURVE WEAK ACID – STRONG BASE 4. pH strong conjugate base

3. pH of salt 2. pH of Buffer

1. pH weak acid

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CALCULATIONS OF TITRATION CURVES How to perform calculations for acid-base titrations? 1. Write the balanced equation for the reaction resulting from the species present in solution. 2. Calculate the moles of reactants. 3. Determine the limiting reactant or product. 4. Calculate the moles of the required reactant or product. 5. Convert to grams or volume if required. The equivalence point in an acid-base titration is defined by the stoichiometry and not by the pH (i.e. 50 ml 0.1M will neutralize 50 ml 0.1 M HCl, and will also neutralize 50 ml 0.1M CH 3COOH). The equivalence point occurs when enough titrant has been added to react exactly with the acid or base being titrated.

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ACID-BASE INDICATORS •

•

•

pH= pKin + log

[]

−

[ ]

For a typical acid-base indicator with dissociation constant Kin, the color changes over a range of pH values given by the pk in 1. A good indicator should have a p K in value that is close to the expected pH at the equivalence point. Strong base (analyte) versus strong acid

PHCM223,SS16, 26-03-2016

Weak base (analyte) versus strong acid


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CHOICE OF INDICATOR IN POLYPROTIC TITRATIONS There are 2 important points: the equivalence point and the mid point. At the midpoint, pH=pKa At the equivalence point, the pH is the average of the 2 pKa values before and after that point

Once you calculate the pH at the equivalence point, you can select the proper indicator that would change its color around that value PHCM223,SS16, 26-03-2016


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PROBLEM 1 What percent of formic acid (HCOOH) is dissociated in a 0.1 M solution of formic acid? The K a of formic acid is 1.77 x 10-4.

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PROBLEM 2 If the H+ concentration is 0.00001 M, what is the OHconcentration? since pH = - log [H+]= -log (1.0 x 10-5) pH = -(-5) = 5 pH + pOH = 14 pOH = 14 - 5 = 9

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PROBLEM 3 pKa for phenolphthalein is 9.3 at 25°C. a) Calculate ratio of its anionic form to acid form at pH 8.2 and at pH 10. b) Using these values, explain the color change within this pH range. a) At pH = 8.2:

8.2 = 9.3 + log (base form / acid form) log (base form / acid form) = -1.1  ratio of base form to acid form = 0.0794 to 1 (≈ 8 to 100) a) At pH = 10.:

10 = 9.3 + log (base form / acid form)  log (base form / acid form) = 0.7 ratio of base form to acid form = 5.01 to 1 ( ≈ 500 to 100) b) the anionic (or base) form that is colored pink. The acid form is colorless. At pH = 8.3, the pink form is in the minority at this pH. For every 100 acid (colorless) forms, there are only 8 base (pink) forms. At pH = 10, the colorless form is in the minority. For every 100 acid (colorless) forms present, there are now 500 base (pink) forms present.

This means that from pH = 8.3 to pH = 10., there has been increase in pink forms (from 8:100 to 500:100). While there might be a slight pink color pH 8.3, the population of pink forms has greatly increased by pH = 10., to the point where the pink color is easily seen. PHCM223,SS16, 26-03-2016


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PROBLEM 4 Aspirin has a pKa of 3.4. What is the ratio of A¯ to HA in: a) blood (pH = 7.4) b) the stomach (pH = 1.4) a) pH = pKa + log [A¯] / [HA] 7.4 = 3.4 + log [A¯]/ [HA] 4 = log [A¯]/ [HA] 104 = [A¯]/ [HA]= 10000 b) 1.4 = 3.4 + log [A¯]/ [HA]

- 2.0 = log [A¯]/ [HA] 10-2 = [A¯]/ [HA] = 0.01 PHCM223,SS16, 26-03-2016


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PROBLEM 5 What is the pH when 25.0 mL of 0.200 M of CH 3COOH has been titrated with 35.0 mL of 0.100 M NaOH? Acetic acid pK a=4.752 CH3COOH + NaOH  CH3COONa + H2O Step 1: Determine moles of acetic acid and NaOH before mixing:

CH3COOH: (0.200 mol/L) (0.0250 L) = 0.00500 mol NaOH: (0.100 mol/L) (0.0350 L) = 0.0035 mol Step 2 : Determine moles of acetic acid and sodium acetate after mixing:

Remaining CH3COOH: 0.00500 mol - 0.00350 mol = 0.00150 mol Newly formed CH3COONa: 0.0035 mol Step 3: Use the Henderson-Hasselbalch Equation:

pH = 4.752 + log [(0.00350 mol/0.060 L) / (0.0015 mol/0.060 L)] pH = 4.752 + log 2.333 pH = 4.752 + 0.368 pH = 5.120 PHCM223,SS16, 26-03-2016


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PROBLEM 6 Determine the pH of a 0.10 M solution of H2SO4. Considering the total [H+] ions obtained from the first and the second dissociation (K a2 = 1.3 x 10-2 ) H2SO4 has two protons, so it is a diprotic acid that undergoes two sequential ionizations in water: First ionization: H2SO4(aq) → H+(aq) + HSO4-(aq) Second ionization: HSO4-(aq) ⇔ H+(aq) + SO42-(aq) Sulfuric acid is a strong acid, so its first dissociation approaches 100%. While for the second ionization HSO4- is a weak acid, so the H + is in equilibrium with its conjugate base. Ka2 = [H+][SO42-]/[HSO4-] = 1.3 x 10 -2 Ka2 = (0.10 + x)(x)/(0.10 - x) Since x can not be ignored (as requested by the problem), it's necessary to use the quadratic formula to solve for x: x2 + 0.11x - 0.0013 = 0

x = 1.1 x 10 -2 M

The sum of the first and second ionizations gives the total [H +] at equilibrium. 0.10 + 0.011 = 0.11 M pH = -log[H+] = 0.96 (If only the first dissociation would have been considered, then pH= -log 0.1= 1 PHCM223,SS16, 26-03-2016


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PROBLEM 7 A solution containing 50.00 mL of 0.1800 M NH3 (Kb = 1.77 x 10-5) is being titrated with 0.3600 M HCl. Calculate the pH: a. initially b. After the addition of 5.00 mL of HCl c. After the addition of a total volume of 12.50 mL HCl d. After the addition of a total volume of 25.00 mL of HCl e. After the addition of 26.00 mL of HCl NH3 + HCl  NH4Cl

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PROBLEM 7 b): 1) Determine moles of NH 3 and HCl before mixing:

NH3: (0.1800 mol/L) (0.0500 L) = 0.00900 mol HCl: (0.3600 mol/L) (0.0050 L) = 0.00180 mol 2) Determine moles of ammonia and ammonium ion after mixing: NH3: 0.00900 mol - 0.00180 mol = 0.00720 mol NH 3 remaining NH4+: 0.00180 mol produced 3) Use the H-H Equation: pH = pKa + log (base/acid) pH = 9.248 + log [(0.0072/0.055) / (0.0018/0.055)] pH = 9.850

```` Note that the pKa was used, NOT the pKb. (pKa = pKw – pKb )

c): 1) Determine moles of NH 3 and HCl before mixing: NH3: (0.1800 mol/L) (0.0500 L) = 0.00900 mol HCl: (0.3600 mol/L) (0.01250 L) = 0.00450 mol 2) Determine moles of ammonia and ammonium ion after mixing: NH3: 0.00900 mol - 0.00450 mol = 0.00450 mol NH 3 remaining NH4+: 0.00450 mol produced 3) Use the H-H Equation: pH = pKa + log (base/acid) pH = 9.248 + log (0.0045 / 0.0045) pH = 9.248 PHCM223,SS16, 26-03-2016


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PROBLEM 7 Solution to part (d):

1) Determine moles of NH 3 and HCl before mixing: NH3: (0.1800 mol/L) (0.0500 L) = 0.00900 mol HCl: (0.3600 mol/L) (0.02500 L) = 0.00900 mol 2) Determine moles of ammonia and ammonium ion after mixing: NH3: 0.00900 mol - 0.00900 mol = zero mol NH 3 remaining NH4+: 0.00900 mol produced This solution is no longer a buffer. It is now the salt of a weak base and its solution will be acidic. The H-H Equation is not used to determine the pH of the solution. Use calculation of pH of salts. The salt is NH 4Cl (see lecture 2, slide 10, solution pH = 5.084).

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PROBLEM 7 e)This solution is now a solution of a strong acid (the presence of the weak acid NH 4+ may be ignored).

The new molarity of the HCl: M1V1 = M2V2 (0.36) (1.00 mL) = (x) (76.00 mL) x = 0.0047 M = [H +] pH = - log [H +] pH = 2.324

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REFERENCES •

•

•

•

http://www.chemteam.info/AcidBase/Hendersonhasselbalch-problems.html http://chemistry.about.com/od/acidbaseproblems/a/Polyprot ic-Acid-Example-Chemistry-Problem.htm http://www.sparknotes.com/chemistry/acidsbases/phcalc/pr oblems.html http://www2.tku.edu.tw/~tscx/2009chinese/other/tb14.pdf

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PHCM223 Midterm Revision SS16 443

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