Review of Acids. HClStrong Acid HCl  H + + Cl -0.10 M pH = -Log[H + ] =1.0 No ICE HFWeak Acid HF ⇌ H + + F - I0.10 M00 CxxxCxxx E0.10 - xxx small Ka.

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Presentation transcript:

Review of Acids

HClStrong Acid HCl  H + + Cl M pH = -Log[H + ] =1.0 No ICE HFWeak Acid HF ⇌ H + + F - I0.10 M00 CxxxCxxx E xxx small Ka x 2 =3.5 x x = M 0.10 pH = -Log[ ] =2.23

NaOHStrong Base NaOH  Na + + OH M pOH = -Log[OH - ] =0.70 No ICE NH 3 Weak Base NH 3 + H 2 O ⇌ NH OH - I0.20 M 0 0 Cx x x E x x x small Kb x 2 =Kb = Kw= 1.0 x = x Ka 5.6 x x = M pOH = -Log[ ] =2.73 pH=11.27

CaO Basic Anhydride CaO + H 2 O  Ca(OH) 2 SO 2 Acid Anhydride SO 2 + H 2 O  H 2 SO 3 Salts NaClneutral salt NaCl  Na + + Cl - KCNbasic salt CN - + H 2 O ⇌ HCN + OH - hydrolysis NH 4 Clacidic salt NH H 2 O ⇌ H 3 O + + NH 3 hydrolysis

FeCl 3 acidic salt Fe(H 2 O) 6 3+ ⇌ H + + Fe(H 2 O) 5 (OH) 2+ hydrolysis- copy off chart! Know the pH scale HClHFNaClNH 3 NaOH Assuming 1M

BuffersMaintain the pH Weak Acid + Salt ( conjugate) HFKF HF ⇌ H + + F - HighLowHigh IndicatorChange colour with pH HInd ⇌ H + + Ind - AcidLowBasic

IndicatorsChange colour with pH methyl red HInd ⇌ H + + Ind - redyellow page trans pt[HInd] = [Ind - ] colour is orange and Ka = [H + ] pH = =5.4 2 Ka = [H + ] = = 4 x 10 -6

Acid RainSO 2 NO 2 pH  5 Normal RainCO 2 pH = 6 AmphriproticH 2 C 2 O 4 acid HC 2 O 4 - ampriprotic C 2 O 4 2- base Is HC 2 O 4 - an acid or base? Ka (HC 2 O 4 - ) =6.4 x Kb (HC 2 O 4 - ) =1.0 x =1.7 x x acid

Is NH 4 CH 3 COO an acid or base? Ka (NH 4 + ) =5.6 x Kb (CH 3 COO - ) =1.0 x = 5.6 x x neutral

100.0 mL M H 2 SO 4 is mixed with mL of M NaOH. What is the pH of the resulting solution? H 2 SO 4 + 2NaOH L x mole L x mole 1 L I mole mole C mole mole E0 mole mole [ NaOH ]= mole= M L pOH = 2.20 pH = 11.80

Complete the reaction and state if the reactants or products are favoured. HCO HSO 3 - BaseAcid HCO HSO 3 - ⇌ H 2 CO 3 + SO 3 2- weakerweaker stronger stronger baseacidacidbase Reactants are favoured

Know the Equations E + H 2 O ⇌ H + + OH - Kw = [H + ][OH - ]=1.0 x pH = -Log[H + ]pOH = -Log[OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH pH + pOH = pKw = 25 o other temperatures pure water is always neutral & [H + ] = [OH - ] & pH = pOH  7