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Calculating pH of strong acids and bases

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Strong acids or bases are those which dissociate completely. HCl(aq) + H 2 O(l) → H 3 O + (aq) + Cl - (aq) So in a 0.5 mol L -1 solution of HCl there will be 0.5 mol L -1 of H 3 O + and 0.5 mol L -1 of Cl -. Similarly, NaOH is a strong base because 0.1 mol L -1 of NaOH(aq) contains 0.1 mol L -1 of Na + and 0.1 mol L -1 of OH -. There are only four strong acids that we normally deal with: HCl, HBr, HNO 3 and H 2 SO 4. All other acids should be considered as weak acids unless you are told otherwise. Likewise, the only strong bases you are likely to meet are NaOH, KOH and Ca(OH) 2.

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Weak acids or bases differ from strong ones in that they only partially dissociate. Their reactions with water are equilibrium reactions. Ethanoic acid (in vinegar) is a weak acid. Most ethanoic acid molecules remain as molecules in water: only a few react to form CH 3 COO - and H 3 O +. Later on in this unit you will find out how to calculate exactly how much that ‘most’ is.

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Example 1: Calculating the pH of a strong acid. Calculate the pH of a 0.0672 mol L -1 solution of nitric acid. Since nitric acid is a strong acid: HNO 3 (aq) + H 2 O(l) → H 3 O + (aq) + NO 3 - (aq) c (HNO 3 ) = 0.0672 mol L -1 so [H 3 O + ] = 0.0672 mol L -1 pH = -log[H 3 O + ] = -log(0.0672) = 1.17 Remember that [x] means ‘the concentration of x’. Do this problem on your calculator and check that you can get this answer.

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Example 2: Calculating the pH of a strong acid. Calculate the pH of a 2.41 × 10 -3 mol L -1 solution of sulfuric acid. Each molecule of H 2 SO 4 will donate two protons: H 2 SO 4 (aq) + H 2 O(l) → 2H 3 O + (aq) + SO 4 2- (aq) Therefore c (H 2 SO 4 ) = 2.41 × 10 -3 mol L -1 But [H 3 O + ] = 2 × 2.41 × 10 -3 mol L -1 = 4.82 × 10 -3 mol L -1 pH = -log[H 3 O + ] = -log(4.82 × 10 -3 ) = 2.32 Do this problem on your calculator and check that you can get this answer.

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Example 3: Calculating the pH of a strong base. Calculate the pH of a 0.00250 mol L -1 solution of potassium hydroxide. KOH is a strong base, therefore c (KOH) = 0.00250 mol L -1 = [OH - ] Before we can calculate the pH we need to know [H 3 O + ]. In any aqueous solution at room temperature [H 3 O + ][OH - ] = 10 -14 Therefore

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So for our 0.00250 mol L -1 solution of potassium hydroxide: [OH - ] = 0.00250 mol L -1

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An alternative method for calculating pH from [OH – ] This method is shorter than converting [OH – ] to [H 3 O + ], but may be harder to understand. Just as we can write pH = –log[H 3 O + ] so we can write pOH = –log[OH – ] And since 1 10 –14 = [H 3 O + ][OH – ] –log(1 10 –14 ) = (–log[H 3 O + ]) + (–log[OH – ]) 14 = pH + pOH If you don’t understand the mathematics of logs, don’t worry – just learn the bottom formula.

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Calculate the pH of a 0.00250 mol L -1 solution of potassium hydroxide.

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