18.4 Acid-Base Titrations. Assessment Statement 18.4.1 Sketch the general shapes of graphs of pH against volume for titrations involving strong and weak.

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Presentation transcript:

18.4 Acid-Base Titrations

Assessment Statement Sketch the general shapes of graphs of pH against volume for titrations involving strong and weak acids and bases, and explain their important features.

Starter Calculate the pH of 0.1 moldm -3 HCl

Strong Acid –Strong Base 50 cm 3 of a 0.1 moldm -3 HCl and titrating with 01M NaOH [H + ] = 1 x10 -1 therefore the PH will be 1.

After 49 cm 3 NaOH has been added…… 1 cm 3 of HCl will be left in the 99ml (50 cm 3 HCL and 49 cm 3 NaOH) Therefore 1 cm 3 /99 cm 3 x 0.10M = M Hence [H + ] = 1 x10 -3 so pH = -log [H + ] pH = -log [1 x10 -3 ] pH = 3

At 50 cm 3 NaOH After 49.1 ml NaOH has been added…… 0.9 cm 3 of HCl will be left in the 99.1 cm 3 (50 cm 3 HCL and 49.1 cm 3 NaOH) Therefore 0.9 cm 3 /99.1 cm 3 x 0.10 moldm -3 = 9.0 x10 -4 moldm -3 Hence [H + ] = 9 x10 -4 so pH = -log [H + ] pH = -log [9 x10 -4 ] pH = 3.04

Your Turn Try calculating the pH after adding another increments of 0.1 cm 3 Use Excel to create a spreadsheet by adding 100 cm 3 NaOH at 0.1 cm 3 increments. Construct a graph of volume NaOH Vs pH

Equivalence Point After EQUIVALENCE POINT has been reached and the solution is neutral. All the [H + ] from the acid has been neutralised and the pH will be 7 (just made up with dissociation from water molecules).

After Neutralisation At 51 cm 3 of NaOH 1 cm 3 of NaOH is present in 101 cm 3 Therefore 1 cm 3 /101 cm 3 x 0.10 moldm -3 = 9.0 x10 -4 moldm -3 Hence [0H - ] = 9 x10 -4 so p0H = -log [0H - ] p0H = -log [9 x10 -4 ] p0H = 3.00 Therefore the pH = = 11

Characteristics of the Graph Very rapid pH at the equivalence point Any common PH indicator can be used as all change colour in this pH range.

Weak Acid – Strong Base 50 cm 3 of ethanoic acid is titrated with 0.1 moldm -3 NaOH Ka(CH 3 COOH) = 1.8 x 10 -5

Initial pH K a = ( [H+][A-]) / [HA] K a = ( [H+][CH 3 COO -1 ]) / [CH 3 COOH] 1.8 x = [H+] 2 / 0.1 moldm = [H+] so pH = -log [H + ] pH = -log [0.0013] pH = 3.04

Use Excel to create a spreadsheet by adding 100 cm 3 NaOH at 0.1 cm 3 increments. Construct a graph of volume NaOH Vs pH

Adding NaOH When 25 cm 3 of NaOH has been added, we will have an IDEAL BUFFER SOLUTION (half the equivalence point has been reached. pH is equal to the pKa so pH = -log [H + ] pH = -log [1.8 x ] pH = 4.76

At 49 cm 3 NaOH, the Concentration of CH 3 COO - is approximately halved to 0.05moldm -. See if you can calculate the pH (ans 6.44)

Equivalence and Salt Hydrolysis The pH at the equivalence point is not 7 due to SALT HYDROLYSIS. When salts made from weak acids and strong bases the ions absorb hydrogen ions to form mainly an undissociated salt leaving excess hydroxide ions.

Characteristics of the graph After the equivalence point the graph will follow same pattern as strong acid – strong base (as only now adding NaOH The end point is much more narrow pH range. Suitable indicator would be phenolphthalein which changes colour in pH range 8.2 – 10.0.

Strong acid – Weak Base Characteristics of the graph are similar to strong acid – strong base BUT The inflection point is shorter as it is a weak base and doesn’t achieve high pH values. You can perform the calculations as before add determine the pH when 50.1 ml of acid has been reached.

Weak acid – Weak Base You can perform the calculations as before add determine the pH when 50.1 ml of acid has been reached. For IB you need to be able to sketch, describe and explain these graphs.