Addition of Acid or Base to Solution CE 541

When adding Base or Acid to Solution: Interaction between different chemical species occur Interaction between different chemical species occur Chemical equilibrium established Chemical equilibrium established Nature of changes depends on Nature of changes depends on Nature of species Nature of species Concentration of species Concentration of species Changes can be predicted by solving equilibrium problem Changes can be predicted by solving equilibrium problem It is cumbersome It is cumbersome Acid-base titration can be used to help understand the process Acid-base titration can be used to help understand the process What is Titration? It is a procedure by which a measured amount of a chemical is added to solution in order to bring about the desired change.

Titration of Strong Acids and Bases Strong Base Titrated with Strong Acid High initial pH (12-13) High initial pH (12-13) Slowly declines to pH ≈ 10 Slowly declines to pH ≈ 10 Rapid decline to pH ≈ 4 Rapid decline to pH ≈ 4 Slow decline Slow decline Between pH 10 – 4 curve is vertical Between pH 10 – 4 curve is vertical Equivalence point between pH 4 and 10 Equivalence point between pH 4 and 10 Equivalence point is reached when equivalents of base (or acid) added equal to equivalents of base (or acid) initially present

Strong Acid Titrated with Strong Base

Titration Problems Can be solved using charge balance. If strong HCl is titrated with strong NaOH NaOH + HCl → H 2 O + Na + + Cl - Charge balance [Na + ] + [H + ] = [OH - ] + [Cl - ] to get the exact pH, activities must be used. However, concentration can produce close solution.

If C 0 = molar concentration of HCl V 0 = volume of HCl to be titrated C = molar concentration of NaOH V = volume of NaOH added Then:

Re-arrange if C >> C 0, then V can be ignored. Study Example 4.18

Titration of Weak Acids and Bases Weak Acid Titrated with Strong Base Curve character depends on acid Curve character depends on acid Monobasic → yields one H + ion Monobasic → yields one H + ion Polybasic → yields more than one H + ion Polybasic → yields more than one H + ion Initial pH depends on degree of ionization Initial pH depends on degree of ionization Most of natural waters contain weak acids and bases Most of natural waters contain weak acids and bases

Titration of Monobasic Weak Acid with Strong Base Acid ionizes as follows: HA ↔ H + + A - The equilibrium relationships are:

when weak acid is titrated with strong base HA + B + OH - → B + + A - + H 2 O B + is the cation associated with the strong base. The charge balance is [H + ] + [B + ] = [A - ] + [OH - ] (3) [H + ] can be determined by solving equations (1), (2), and (3)

If C 0 = molar concentration of weak acid V 0 = volume of weak acid to be titrated C = molar concentration of strong base V = volume of strong base added Then:

mass balance of strong base cation [B+] = (CV) / (V 0 + V) combining mass balance and equilibrium relationship of weak acid

The charge balance becomes Rearrange The equation can be used to generate titration curve Study Example 19

Beginning of Titration In the initial solution containing the weak acid [B + ] = 0 [B + ] = 0 pH is low  [OH - ] << [H + ] pH is low  [OH - ] << [H + ] So, charge balance becomes [H + ] ≈ [A - ] Originally [H + ] + [B + ] = [A - ] +[OH - ]

Substitute in and solve for [H + ] At the beginning of titration (most of cases) [A] << C 0

so,then, Study Examples 20-21

Midpoint of Titration As titration proceeds HA + B + + OH - → B + + A - + H 2 O [A - ] increases [A - ] increases [HA] decreases [HA] decreases when neutralization reaches 50% [B + ] ≈ (1/2) C 0 [B + ] ≈ (1/2) C 0 [A - ] increases significantly [A - ] increases significantly

so, [H + ] << [B + ] [OH - ] << [A - ] From charge balance ([H + ] + [B + ] = [A - ] + [OH - ]) [B + ] ≈ [A - ] ≈ (1/2) C 0 and since [HA] + [A - ] = C 0 then [HA] ≈ (1/2) C 0

Substitute in

Equivalence Point of Titration At that point, equivalents of base added are equal to equivalents of acid in the original solution (f = 1) So, [B + ] = C 0 The pH value is high enough to assume [H + ] << [OH - ] C 0 ≈ [A - ] + [OH - ] C 0 – [A - ] ≈ [OH - ]

Since C 0 – [A - ] = [HA] Thus, Substitute in Gives

At the end of titration [A - ] ≈ C 0 then,

When titrating weak base with strong acid, pH at various points can be determined using the same approach. Beginning point pH ≈ pK w – (1/2) pK B + (1/2) log C 0 Midpoint pH ≈ pK w – pK B Equivalence point pH ≈ (1/2) (pK w – pK B – logC 0 )

And when titrating weak acid with strong base Beginning of titration pH ≈ (1/2)(pK A – log C 0 ) Midpoint pH ≈ pK A Equivalence point pH ≈ (1/2) (log C 0 + pK A + pK w ) Study Examples 22 and 23

Calculate the pH of the equivalence point in the titration of solutions containing the following concentrations of sodium bicarbonate with sulfuric acid: (a) 10 mg/l; (b) 100 mg/l; © 1000 mg/l

A 500ml solution contains 100 mg of NaOH. Calculate the following 1. initial pH of the solution 2. pH after addition of 2 ml of 1 N H 2 SO 4 3. pH after addition of 4 ml of 1 N H 2 SO 4 4. ml of 1 N H 2 SO 4 required to reach the equivalence point

Buffers are substances in solution that offer resistance to changes in pH as acids or bases are added or formed within the solution are substances in solution that offer resistance to changes in pH as acids or bases are added or formed within the solution buffer solutions usually contain mixtures of weak acids and their salts or weak bases and their acids (conjugates) buffer solutions usually contain mixtures of weak acids and their salts or weak bases and their acids (conjugates)if

Rearrange So, pH of a buffer solution depends on the ratio of salt concentration to acid concentration Study Example 24

Calculate the pH of a buffer solution prepared by mixing 2.4 g of acetic acid (CH 3 COOH) and 0.73 g of sodium acetate (CH 3 COONa) in 1 liter of water.

Buffering Index or Buffering Capacity can be determined by: where β is the buffering capacity or index of weak acid plus its conjugate salt. If we have weak salt (base) plus its conjugate acid, then K A is replaced by (K w / K B ) Study Example 25