9 Aqueous Solutions

Water as a Polar Solvent ,[object Object], +  + H H  -  + O

[object Object],The shape of the molecule is important to its polarity. It must have a positive and negative end. H H  -  +  + O

[object Object],Na + Cl - Na + Na + Cl - Cl - Cl - H H  -  +  + O

[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],C C H H H H H H O  + Polar substances will dissolve in ethanol which has a polar and a non-polar part. Water and ethanol are therefore miscible. Note the ethanol and water do not dissociate into ions!  -  +  +  - H H - O

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],HCl + H 2 O  H 3 O + + Cl - H + .. H H  -  +  + O

Writing Ionic equations ,[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object]

Solubility ,[object Object],[object Object]

Precipitation Reactions ,[object Object],[object Object]

Acid-Base Reactions ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object]

Writing the net ionic equation for an acid base reaction ,[object Object],[object Object],[object Object],[object Object]

Ammonia ,[object Object],[object Object],[object Object]

[object Object],[object Object]

The pH scale ,[object Object],[object Object],[object Object],[object Object]

[object Object],-log 0.01 = 2 What is the pH of a 0.1 molar solution of sulphuric acid? H 2 SO 4  2H + + SO 4 2-  [H + ] = 2 x 0.1 = 0.2 mol/l And pH = -log 0.2 = 0.7

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],b) [H + ] = 3 x 0.05 mol/L pH = -log 0.15  pH = 0.82

Finding pH from [H + ] ,[object Object],Antilog -3 = 0.001  [H] = 0.001 What is the [HCl]? ,[object Object],[object Object]

[object Object],Antilog -2.4 = 0.004 mol/L  [H+] = 0.004 mol/L H 2 SO 4  2H+ + SO 4 2-  [H 2 SO 4 ] = 0.004/2 = 0.002 mol/L What is the [H 2 SO 4 ]?

Finding pH from [OH - ] ,[object Object],[object Object],[object Object],What is the pH of a 0.1 mol/L solution of sodium hydroxide? pOH = -log 0.1 = 1  pH = 14 – 1 = 13

[object Object],[object Object],[object Object],a) pOH = -log 0.001 = 3  pH = 14 -3 = 11 b) [OH - ] = 2 x 0.05  pOH = -log 0.1 = 1  pH = 14 -1 = 13

[object Object],RMM Ca(OH) 2 = 40 + (2 x 16) + (2 x 1) = 74 1.00g = 1.00 = 0.0135mol 74 Moles [OH - ] = 2 x 0.0135 = 0.027 Moles per litre = 2 x 0.027 = 0.054 pOH = -log 0.054 = 1.27 pH = 14 – 1.27 = 12.3

9 Aqueous Solutions

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9 Aqueous Solutions