 # ANALYTICAL CHEMISTRY CHEM 3811 CHAPTER 10 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university.

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ANALYTICAL CHEMISTRY CHEM 3811 CHAPTER 10 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university

CHAPTER 10 ACID-BASE TITRATIONS

STRONG ACID – STRONG BASE - Write balanced chemical equation between titrant and analyte - Calculate composition and pH after each addition of titrant - Construct a graph of pH versus titrant added

- Consider titration of base with acid H + + OH - → H 2 O K = 1/K w = 1/10 -14 = 10 14 - Equilibrium constant = 10 14 - Reaction goes to completion STRONG ACID – STRONG BASE

H + + OH - → H 2 O At equivalence point moles of titrant = moles of analyte (V titrant)(M titrant) = (V analyte)(M analyte) STRONG ACID – STRONG BASE

Consider 50.00 mL of 0.100 M NaOH with 0.100 M HCl - Three regions of titration curve exists - Before the equivalence point where pH is determined by excess OH - in the solution - At the equivalence point where pH is determined by dissociation of water (H + ≈ OH - ) - After the equivalence point where pH is determined by excess H + in the solution STRONG ACID – STRONG BASE

First calculate the volume of HCl needed to reach the equivalence point (V HCl)(0.100 M) = (50.00 mL)(0.100 M) Volume HCl = 50.00 mL STRONG ACID – STRONG BASE

Before the equivalence point Initial amount of analyte (NaOH) = 50.00 mL x 0.100 M = 5.00 mmol After adding 1.00 mL of HCl mmol H + added = mmol OH - consumed mmol H + = (1.00 mL)(0.100 M) = 0.100 mmol mmol OH - remaining = 5.00 – 0.100 = 4.90 mmol STRONG ACID – STRONG BASE

Before the equivalence point Total volume = 50.00 mL + 1.00 mL = 51.00 mL [OH - ] = 4.90 mmol/51.00 mL = 0.0961 M pOH = - log(0.0961) = 1.017 pH = 14.000 - 1.017 = 12.983 STRONG ACID – STRONG BASE

- Repeat calculations for all volumes added - Increments can be large initally but must be reduced just before and just after the equivalence point (around 50.00 mL in this case) - Sudden change in pH occurs near the equivalence point - Greatest slope at the equivalence point STRONG ACID – STRONG BASE

At the equivalence point - pH is determined by the dissociation of water H 2 O ↔ H + + OH - K w = x 2 = 1.0 x 10 -14 x = 1.0 x 10 -7 pH = 7.00 (at 25 o C) STRONG ACID – STRONG BASE

After the equivalence point - Excess H + is present After adding 51.00 mL of HCl Excess HCl present = 51.00 – 50.00 = 1.00 mL Excess H + = (1.00 mL)(0.100 M) = 0.100 mmol Total volume of solution = 50.00 + 51.00 = 101.00 mL [H + ] = 0.100 mmol/101.00 mL = 9.90 x 10 -4 M pH = -log(9.90 x 10 -4 ) = 3.004 STRONG ACID – STRONG BASE

pH Volume of HCl added (mL) 7 50.00 Equivalence point (maximum slope or point of inflection)

pH Volume of NaOH added (mL) 7 50.00 Equivalence point (maximum slope or point of inflection) TITRATION CURVE

Compare titration of HCl with NaOH and H 2 SO 4 with NaOH (same volume and same concentration of acid) HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) H 2 SO 4 (aq) + 2NaOH(aq) → Na 2 SO 4 (aq) + 2H 2 O(l) Net ionic equation in both cases: H + (aq) + OH - (aq) → H 2 O(l) or H 3 O + (aq) + OH - (aq) → 2H 2 O(l) 1 mol HCl conctributes 1 mol H 3 O + 1 mol H 2 SO 4 contributes 2 mol H 3 O + STOICHIOMETRY AND TITRATION CURVE

pH Volume of NaOH added (mL) STOICHIOMETRY AND TITRATION CURVE HCl H 2 SO 4

WEAK ACID – STRONG BASE Consider 50.00 mL 0f 0.0100 M acetic acid with 0.100 M NaOH pK a of acetic acid = 4.76 HC 2 H 3 O 2 + OH - → C 2 H 3 O 2 - + H 2 O For the reverse reaction K b = K w /K a = 5.8 x 10 -10 Equilibrium constant = 1/K b = 1.7 x 10 9 So large that we can assume the reaction goes to completion

WEAK ACID – STRONG BASE Determine volume of base at equivalence point mmol HC 2 H 3 O 2 ≈ mmol OH - (V NaOH)(0.100 M) = (50.00 mL)(0.0100 M) Volume NaOH = 5.00 mL

WEAK ACID – STRONG BASE Four types of calculations to be considered Before OH - is added - pH is determined by equilibrium of weak acid HA ↔ H + + A - x = 4.1 x 10 -4 pH = 3.39

WEAK ACID – STRONG BASE Before equivalence point By adding OH - a buffer solution of HA and A - is formed After adding 0.100 mL OH - HA + OH - → A - + H 2 O Initial mmol 0.500 0.0100 0 Final mmol 0.490 0 0.0100

WEAK ACID – STRONG BASE Before equivalence point

WEAK ACID – STRONG BASE Before equivalence point If volume of OH - added is half the volume at equivalence point HA = A - = 0.250 mmol pH = pK a = 4.76

WEAK ACID – STRONG BASE At equivalence point Volume of OH - = 5.00 mL mmol OH - = (5.00 mL)(0.100 M) = 0.500 mmol HA is used up and [HA] = 0 mmol A - = 0.500 mmol

WEAK ACID – STRONG BASE At equivalence point Only A - is present in solution A - + H 2 O ↔ HA + OH - [A - ] = (0.500 mmol)/(50.00 mL + 5.00 mL) = 0.00909 M

WEAK ACID – STRONG BASE At equivalence point K b = K w /K a = 5.8 x 10 -10 x = [OH - ] = 2.3 x 10 -6 pOH = 5.64 pH = 14.00 – 5.64 = 8.36 pH is not 7.00 but greater than 7.00 (pH at equivalence point increases with decreasing strength of acid)

WEAK ACID – STRONG BASE After equivalence point - Strong base (OH - ) being added to weak base (A - ) - pH is determined by the excess [OH - ] (approximation) - After adding 5.10 mL OH - [OH - ] = (0.10 mL)(0.100 M)/(50.00 mL + 5.10 mL) = 1.81 x 10 -4 pH = 14.00 – pOH = 14.00 – 3.74 = 10.26

WEAK ACID – STRONG BASE pH Volume of NaOH added (mL) 8.36 5.00 Equivalence point (maximum slope or point of inflection) pH = pK a Minimum slope pK a 2.50

STRONG ACID – WEAK BASE - The reverse of weak acid and strong base B + H + → BH + - Similarly assume reaction goes to completion

Consider 50.00 mL of 0.0100 M pyridine with 0.100 M HCl K b of pyridine = 1.6 x 10 -9 Determine volume of acid at equivalence point mmol pyridine ≈ mmol H + (V HCl)(0.100 M) = (50.00 mL)(0.010 M) Volume HCl = 5.00 mL STRONG ACID – WEAK BASE

Four types of calculations to be considered Before H + is added - pH is determined by equilibrium of weak base (determined using K b ) B + H 2 O ↔ BH + + OH - STRONG ACID – WEAK BASE

Before equivalence point - By adding H + a buffer solution of B and BH + is formed STRONG ACID – WEAK BASE - When volume of H + added = half the volume at equivalent point pH = pK a (for BH + )

At equivalence point - B has been converted into BH + - B is used up and [B] = 0 pH is calculated by considering BH + BH + ↔ B + H + pH is not 7.00 but less than 7.00 STRONG ACID – WEAK BASE

After equivalence point - Strong acid (H + ) is being added to weak acid (BH + ) pH is determined by the excess [H + ] (approximation) STRONG ACID – WEAK BASE

pH Volume of HCl added (mL) 5.00 Equivalence point (maximum slope or point of inflection) pH = pK a Minimum slope pK a 2.50

END POINT Use of Indicators - Indicators are acids or bases so a few drops of dilute solutions are used to minimize indicator errors - Acidic color if pH ≤ pK HIn - 1 - Basic color if pH ≥ pK HIn + 1 - A mixture of both colors if pK HIn - 1 ≤ pH ≤ pK HIn + 1 - Use an indicator whose transition range overlaps the steepest part of the titration curve

END POINT Use of pH Electrodes - The end point is where the slope of the curve is greatest - The end point is the volume at which the first derivative of a titration curve is maximum - The end point is the volume at which the second derivative of a titration curve is zero

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