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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 on theme: "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."— Presentation transcript:

1 Review of Acids

2 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 x 2 =3.5 x 10 -4 x =0.005916 M 0.10 pH = -Log[0.005916] =2.23

3 NaOHStrong Base NaOH  Na + + OH - 0.20 M pOH = -Log[OH - ] =0.70 No ICE NH 3 Weak Base NH 3 + H 2 O ⇌ NH 4 + + OH - I0.20 M 0 0 Cx x x E0.20 - x x x small Kb x 2 =Kb = Kw= 1.0 x 10 -14 = 1.786 x 10 -5 0.20 Ka 5.6 x 10 -10 x =0.001890 M pOH = -Log[0.001890] =2.73 pH=11.27

4 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 4 + + H 2 O ⇌ H 3 O + + NH 3 hydrolysis

5 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 0 714 HClHFNaClNH 3 NaOH Assuming 1M

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

7 IndicatorsChange colour with pH methyl red HInd ⇌ H + + Ind - redyellow page 7 @ trans pt[HInd] = [Ind - ] colour is orange and Ka = [H + ] pH =4.8 + 6.0=5.4 2 Ka = [H + ] = 10 -5.4 = 4 x 10 -6

8 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 10 -5 Kb (HC 2 O 4 - ) =1.0 x 10 -14 =1.7 x 10 -13 5.9 x 10 -2 acid

9 Is NH 4 CH 3 COO an acid or base? Ka (NH 4 + ) =5.6 x 10 -10 Kb (CH 3 COO - ) =1.0 x 10 -14 = 5.6 x 10 -10 1.8 x 10 -5 neutral

10 100.0 mL 0.100 M H 2 SO 4 is mixed with 220.0 mL of 0.100 M NaOH. What is the pH of the resulting solution? H 2 SO 4 + 2NaOH 0.1000 L x 0.100 mole0.2200 L x 0.100 mole 1 L I0.0100 mole0.0220 mole C0.0100 mole0.0200 mole E0 mole0.0020 mole [ NaOH ]=0.0020 mole=0.00625 M 0.320 L pOH = 2.20 pH = 11.80

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

12 Know the Equations E + H 2 O ⇌ H + + OH - Kw = [H + ][OH - ]=1.0 x 10 -14 pH = -Log[H + ]pOH = -Log[OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH pH + pOH = pKw = 14 @ 25 o C @ other temperatures pure water is always neutral & [H + ] = [OH - ] & pH = pOH  7

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