UNIT IV Titrations

R EVIEW OF T ITRATION C ALCULATIONS What to do…How to do it… 1. Find moles of standard. n = C x V 2. Find moles of sample. Balanced Equation 3. Find [ ] of sample. C = n/V

T HE E QUIVALENCE P OINT The point at where the actual mole ratio of sample/standard is the same as the coefficient ratio in the balanced equation. Ex.HCl + NaOH H 2 O + NaCl At equivalence point : moles of NaOH = moles of HCl Ex.2HCl + Ba(OH) 2 2H 2 O + BaCl 2 At equivalence point : moles of HCl = 2 x mol Ba(OH) 2

T ITRATION C ALCULATIONS Ex. A solution of HCl of unknown concentration was titrated with M Ba(OH) 2. The equivalence point is reached when mL of Ba(OH) 2 is added to mL of the HCl solution. Find the [HCl] in the original sample. ***If a series of volume readings for different Trials are given, you may have to discard a reading that is more than 0.02 or so mL different from the rest of them. This ability to discard far off volume readings and then to calculate the best average volume will be tested!

T ITRATION C ALCULATIONS Ex M NaOH is used to titrate 3 separate 50.0 mL samples of a solution of H 2 SO 4 of unknown concentration. Use the following data table to calculate the [H 2 SO 4 ] in the original solution. Hebden Textbook Page 158 Questions #94 – 97 Trial 1Trial 2Trial 3 Initial Burette Reading (mL) Final Burette Reading (mL)

I NDICATORS Acid-Base Indicators consist of equilibrium mixtures of: A weak acid and its conjugate base HInd Ind - which are: Different Colours

I NDICATORS

Ex.) An indicator HInd has a yellow acid form (HInd) and a red base form (Ind - ). The equilibrium equation representing this indicator is: HInd + H 2 O H 3 O + + Ind -

I NDICATORS HInd + H 2 O H 3 O + + Ind - If reactants are favoured (equilibrium shifts to the LEFT ), then: [HInd] > [Ind - ] So: yellow is much greater than red and the solution will be YELLOW. If products are favoured (equilibrium shifts to the RIGHT ), then: [Ind - ] > [HInd] So: red is much greater than yellow and the solution will be RED.

I NDICATORS HInd + H 2 O H 3 O + + Ind - If there are equal amounts of reactants and products (equilibrium favours neither reactants nor products), then: [HInd] = [Ind - ] So: there is an equal mixture of yellow and red and the solution will be ORANGE.

I NDICATORS When an acid is added to an indicator (mixture of HInd and Ind - ), it increases [ H 3 O + ] causing the equilibrium: HInd + H 2 O H 3 O + + Ind - To shift LEFT so [HInd] > [Ind - ] and the colour of HInd predominates If you add any strong acid (eg. HCl) to an indicator, it will turn the colour of the ACID FORM (HInd).

I NDICATORS When a base is added to an indicator (mixture of HInd and Ind - ), it decreases [ H 3 O + ] causing the equilibrium: HInd + H 2 O H 3 O + + Ind - To shift RIGHT so [HInd] < [Ind - ] and the colour of Ind - predominates If you add any strong base (eg. NaOH) to an indicator, it will turn the colour of the BASE FORM (Ind - ).

I NDICATORS Question : When a drop of 0.1M HCl is added to the indicator bromcresol green, the colour is yellow. When a drop of 0.10M NaOH is added to the indicator, the colour is blue. What colour is the acid form of bromcresol green (HInd)? __________________ What colour is the base form of bromcresol green (Ind - )? __________________ What would the colour be if [HInd] = [Ind - ] for bromcresol green? __________________

T RANSITION P OINT The Transition Point for an indicator is reached when [HInd] = [Ind - ]. This is where you have equal amounts of the colour of HInd and the colour of Ind -. Looking on the Acid-Base Indicators Table: The two colours on the right side of the table for each indicator lists the colour of the ACID FORM first and then the colour of the BASE FORM.

T RANSITION P OINT So the Acid Form of methyl violet (HInd) is YELLOW and the Base Form of methyl violet (Ind - ) is BLUE. The colour at the TRANSITION POINT of Methyl violet would be ______________ The colour at the TRANSITION POINT of Bromcresol green would be ______________ The colour at the TRANSITION POINT of Indigo carmine would be ______________

T RANSITION P OINT AND K A OF I NDICATOR The equilibrium equation for an indicator (HInd) is: HInd + H 2 O H 3 O + + Ind - So the acid form (HInd) can be thought of as a weak acid. Weak acids, as you know, have a Ka. The Ka expression for the weak acid HInd would be: Ka =[H 3 O + ] [ Ind - ] [HInd]

T RANSITION P OINT AND K A OF I NDICATOR The Transition Point for an indicator is reached when [HInd] = [Ind - ] Since [HInd] = [Ind - ] at the transition point, we can cancel them out in the Ka expression. So AT THE TRANSITION POINT: Ka =[H 3 O + ] [Ind - ] [HInd] AT THE TRANSITION POINT: Ka = [H 3 O + ]

T RANSITION P OINT AND K A OF I NDICATOR By definition: pKa = -log Ka So, AT THE TRANSITION POINT: Ka = [H 3 O + ] Take the (-) log of both sides: -log Ka = -log[H 3 O + ] pKa = pH

T RANSITION P OINT AND K A OF I NDICATOR Summary: AT THE TRANSITION POINT: [HInd] = [Ind - ] Ka (indicator) = [H 3 O + ] pKa = pH The colour is a 50/50 mixture of the acid and base colours.

T RANSITION P OINT & T RANSITION R ANGE If you look at the Indicator Table on the back of the Acid Table, there is a column entitled pH Range in which Colour Change Occurs. As the pH is gradually raised, the colour does not instantaneously change from acid colour to base colour. There is a gradual change over a range of pHs.

T RANSITION P OINT & T RANSITION R ANGE Ex. Methyl Violet gradually changes from yellow to blue in the pH range of 0.0 – 1.6. When pH is at or below 0.0, the colour of methyl violet is yellow. When pH is 1.6 or above, the colour of methyl violet is blue. But what about between? Between pH of 0.0 and 1.6, there is a mixture of the yellow and the blue form of methyl violet, so the colour is GREEN. (We can refine it even further by saying that between pH of 0.0 and 0.8, the colour is more of a yellow green and between pH 0.8 and 1.6, it is more of a blue green. At a pH of 0.8 (half-way between 0.0 and 1.6), the colour would be simply green!)

T RANSITION P OINT & T RANSITION R ANGE pHThymol BlueOrange IV

F INDING THE K A OF AN I NDICATOR 1. Find the pH at Transition Point: Look on the Indicator Table. Find the midpoint of the pH range by adding the two numbers and dividing by two. 2. pH at the Transition Point = pKa of indicator 3.Ka of indicator = antilog (-pKa) (Since pKa = -log Ka)

F INDING THE K A OF AN I NDICATOR Ex. Find the Ka of Phenol Red. Ex. Find the Ka of Alizarin Yellow.

I NDICATOR C OLOUR A CCORDING TO P H Thymol Blue appears twice on the Indicator Table. H 2 Tb + H 2 O H 3 O + + HTb - HTb - + H 2 O H 3 O + + Tb 2- pHForm(s) which predominate(s) (H 2 Tb, HTb - or Tb 2- ) Approximate Colour & are equal & are equal 10.0

I NDICATOR C OLOUR A CCORDING TO P H Ex. Find the pH and the colour of the given indicators in the following solutions (assume T = 25 o C): SolutionpHColour in Thymol Blue Colour in Methyl Red Colour in Alizarin Yellow 0.2 M HCl 0.01 M HCl M HCl Pure water M NaOH 0.2 M NaOH

I NDICATOR C OLOUR A CCORDING TO P H A variety of indicators can also be used to narrow the known pH range for a solution and help identify the solution: IndicatorColour of SolutionApproximate pH Range Bromthymol blueBlue Thymol blueYellow PhenolphthaleinColourless Approximate pH range of the solution using all information:

I NDICATOR C OLOUR A CCORDING TO P H IndicatorColour of SolutionApproximate pH Range Orange IVYellow Methyl redRed Methyl OrangeRed Approximate pH range of the solution using all information: IndicatorColour of SolutionApproximate pH Range Methyl OrangeYellow Alizarin YellowYellow Thymol BlueGreen Approximate pH range of the solution using all information:

U NIVERSAL I NDICATORS Universal Indicators: Give a variety of colours over a larger pH range. If several indicators are mixed, the combinations of colours can lead to many different colours as we move from one pH to another. Study the 3 tables given on page 162 of your textbook to give you an idea of how universal indicators can be made. The second table is somewhat simplified as it does not include the colours of indicators in their transition ranges. The third table is more precise. Hebden Textbook Pages Questions # ,

U SING I NDICATORS TO R ANK W EAK A CIDS IN O RDER OF S TRENGTHS Ex. An indicator HInd is RED in 0.1M HCl and BLUE in 0.1M NaOH. Give the equilibrium equation for this indicator and write the colour of each form (HInd) and (Ind - ) underneath it: A few drops of this indicator is added to a weak acid called HA 1 and the colour is blue. Which is the stronger acid, HA 1 or HInd?

U SING I NDICATORS TO R ANK W EAK A CIDS IN O RDER OF S TRENGTHS To find out, we write an equilibrium equation (NOT with H 2 O this time!) For reactants, we use the weak acid HA 1 and the base form of the indicator, Ind -. Two acids are not written on the same side of equilibrium equations! HA 1 + Ind - +

U SING I NDICATORS TO R ANK W EAK A CIDS IN O RDER OF S TRENGTHS Since the colour of the indicator was blue, it means that the form of the indicator (HInd or Ind - ) is predominating (favoured by the equilibrium). So the ( reactants/products ) of the equation above are favoured, meaning ( HA 1 /HInd ) is the Weaker Acid or ( HA 1 /HInd ) is the Stronger Acid.

U SING I NDICATORS TO R ANK W EAK A CIDS IN O RDER OF S TRENGTHS Rewrite the equilibrium equation for the indicator and write the colour of each form (HInd) and (Ind - ) underneath it: Next, A few drops of this indicator is added to a weak acid called HA 2 and the colour is red. Which is the stronger acid, HA 2 or HInd? HA 2 + Ind - +

U SING I NDICATORS TO R ANK W EAK A CIDS IN O RDER OF S TRENGTHS Since the colour of the indicator was red, it means that the form of the indicator (HInd or Ind - ) is predominating (favoured by the equilibrium). So the ( reactants/products ) of the equation above are favoured, meaning ( HA 2 /HInd ) is the Weaker Acid or ( HA 2 /HInd ) is the Stronger Acid.

U SING I NDICATORS TO R ANK W EAK A CIDS IN O RDER OF S TRENGTHS So, to summarize the results of both experiments: Experiment 1: __________> __________ Experiment 2: __________ > __________ So, in comparing strengths of HA 1 and HA 2, we can say that __________ > __________

P RACTICAL A SPECTS OF T ITRATION Remember, standard solutions are solutions of accurately known concentration. They are used to titrate solutions of unknown concentration (sample solutions). Note that solid NaOH cannot be used to prepare a standard solution (by weighing it and dissolving it in a known volume of water). NaOH cannot be weighed accurately as it absorbs water and CO 2 from the air as its being weighed ( hygroscopic ).

P RACTICAL A SPECTS OF T ITRATION There are 2 ways to prepare a standard solution accurately: 1. Use a Primary Standard A Primary Standard has the following characteristics: It is obtained in pure and stable form & dissolves completely It does NOT absorb H 2 O or CO 2 from the air (non-hygroscopic) It has an accurately known molar mass It reacts quickly and completely with the sample

P RACTICAL A SPECTS OF T ITRATION An accurately measured mass of the primary standard is weighed and dissolved in an accurately measured volume of water to obtain a solution of accurately known concentration (Standard Solution). Ex g of potassium hydrogen phthalate (KHC 8 H 4 O 4 ) is weighed out and dissolved in enough distilled water to make L of solution. Find the [KHC 8 H 4 O 4 ].

P RACTICAL A SPECTS OF T ITRATION Some Primary Standards are: Na 2 CO 3 (sodium carbonate) KHC 8 H 4 O 4 (potassium hydrogen phthalate) C 6 H 5 COOH (benzoic acid) NEVER NaOH remember, it is highly hygroscopic!

P RACTICAL A SPECTS OF T ITRATION 2. Standardizing a Solution This is done by titrating a solution with a primary standard in order to find its accurate concentration. The standardized solution can then be used to titrate other solutions. Ex. It takes 4.02 mL of M KHC 8 H 4 O 4 to titrate mL of a solution of NaOH. Find the [NaOH]. KHC 8 H 4 O 4 + NaOH H 2 O + KNaC 8 H 4 O 4

P RACTICAL A SPECTS OF T ITRATION This standardized NaOH solution can now be used to titrate other acids of unknown concentration: Ex. It takes mL of standardized M NaOH to titrate a mL sample of an H 2 SO 4 solution. Hebden Textbook Page 165 Questions #121 – 123