# Acids and Bases

## Bronsted-Lowry

ACID: Proton donor

BASE: Proton Acceptor

Acid-Base equilibria involves the transfer of protons.

Dissociation in water                                                                                                                STRONG ACIDS (HCl): Completely dissociate in water                                                          STRONG BASES (NaOH): Ionise completely in water                                                                WEAK ACIDS (CH3COOH): Partially dissociate in water                                                            WEAK BASES (NH3): Partially dissociate in water

Ionic Product of Water

Kw = [H+][OH-]            H2O                  H+  + OH-

Kc =  [H+] [OH-]                                                                                                                                      .       [H2O]                                                                                                                                  Water only dissociates a small amount                                Kw = 1.00 x 10^-14

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## pH Calculations

pH = -Log10[H+]        [  ] ---> concentration in mol dm-3

[H+] from pH                                                                                                                                          If you have pH and want to know [H+]         [H+] = 10 ^-pH

pH of strong monoprotic acids                                                                                                Monoprotic --> each molecule of an acid releases on proton when it dissociates.                          HCl ---> H+ + Cl-      e.g. calculate pH of 0.100 moldm-3 HCl :   -Log(0.0100) = 1.00

pH of a strong diprotic acid                                                                                                         Diprotic--> each molecule of an acid releases 2 protons when it dissociates.                            H2SO4 ---> 2H+ + SO4^^2-  e.g. calculate pH of 1.00 mol dm-3 H2SO4  0.100 x 2  -Log(0.200)=0.7

pH of strong bases                                                                                                                        NaOH and KOH ---> donates 1 OH- per mole of base                                                                      Kw=[H+][OH-]                      1.Find the value of Kw and [OH-], you may be told.                                2. Rearrange the equation ---> [H+] == Kw / [OH-]       3. Know [H+] so do -Log[H+] to find pH

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## Acid Dissociation Constant

HA      H+  +  A-                                                      WEAK ACIDS

Ka = [H+] [A-]                                                                                                                                            [HA]

Assume that [H+] is squared

pH: -Write Ka   -Rearrange to find [H+]2   -take the square root of the value for [H+]2 then -Log[H+]

Concentration:  -Find [H+] by  10^-pH   -Write the Ka expression and then rearrange to find the concentration of the acid ---> [H+]2 / Ka   -Substitute in the values and solve

pKa                                                                                                                                                   pKa = -Log10[Ka]                                     Ka = 10^-pKa

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## Titrations and pH Curves

STRONG ACID + STRONG BASE:

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STRONG ACID + WEAK BASE:

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WEAK ACID + STRONG BASE:

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WEAK ACIDWEAK BASE:

INDICATORS: Pick one changes colour over narrow pH.   Methyl-Orange and Phenolphthalein

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## Titration Calculations

MONOPROTIC ACIDS: -Write out balancd equation  -Decide what yyou know & what you need to know.   -One reagent you know the conc. and vol. ---> calculate the moles of this :                    Moles = conc. (moldm-3) x ( Col. (cm3) / 1000) )                                                                           --Use molar ratios in balanced equations to find how many moes of the other reagent reacted.   -Calculate the know conc. using:  conc. = (mol x 1000) / Vol. (cm3)

DIPROTIC ACIDS: e.g. ethandioic acid.   2 stages to a pH curve with 2 equivalance points

You calculate the same way as monoprotic, BUT you need 2x's as many moles of base as moles of acid.

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## Buffer Action

BUFFER: Resists changes in pH when small amounts of acid or alkali are added.

ACIDIC BUFFER         pH < 7                  Made by mixing weak acid with one of its salts.

Acid Added;   -[H+] Increases -The rise in [H+] should decrease the pH but... Excess H= combines with A- and the equilibrium shifts left.   -[H+] decreases to close to its original value --> pH same.                                                                                                                                              Base Added; -[OH] increases.   - OH- combines with H+ so [H+] decreases.     -the equilibrium shifts right to replace the lost H+     -[H+] increases to close to its original value. --> pH stays almost same.

BASIC BUFFER           pH > 7                  Made by mixing weak base with one of its salts

Acid Added;  -[H+] Increases  -H+ combines with OH- so [OH-] decreases      -Equilibrium shifts right to replace lost OH-.      -[OH-] increases to close to its original value. --> pH stays almost the same.                                                                                                                                         Base Added;  -[OH-] increases  -Excess OH- combines with BH+ and equilibrium shifts left.      -[OH-] decreases to close to its original value.

APPLICATIONS: Shampoo, Biological Washing Powder and Blood

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## Calculation pH of Buffers

Known Concentrations

1. Write out Ka of Weak Acid                                                                                                               2. Rearrange to give expression for [H+]                                                                                             3. Sub in the values.                                                                                                                            4. Solve to find [H+]                                                                                                                                5. -Log[H+]

Another Way

1. Write out equation for neutralisation reaction    Acid + Base  --->  Salt + Water                              2.Calculate moles of acid + base at start using Vol. and Conc.   Mol = Conc. x (Vol. / 1000)             3.Use molar ratios to work which is in excess                                                                                      4. Calculate Conc. of acid and salt in buffer solution by (Mols x 1000) / Vol.            -Work out pH

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