10.0 Acid-Base Buffer Solutions
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Definition: A buffer solution resists changes in pH when small amounts of acid or base is added to it.
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A buffer solution is a mixture of an acid and an alkali. The acid neutralises the alkali added and the alkali neutralises the acid added.
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The mixture cannot be a mixture of a strong acid and a strong alkali, or the two will react with each other (e.g. a mixture of NaOH and HCl would react with each other and thus not behave as an effective buffer).
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If the acid and alkali in the buffer are too weak, however, they will not react effectively with the acid or alkali that are added.
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A suitable mixture is one which contains a mixture of acid and alkali strong enough to react with H3O+ and OH-, but weak enough not to react with each other.
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An ideal mixture for this purpose is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. For example, CH3COOH and CH3COONa, or NH4Cl and NH3. The acids and bases in these mixtures will react with OH- and H3O+ respectively but not with each other.
10.1 Acidic Buffers
10.2 Alkaline Buffers
10.3 Calculating the pH of Buffer Solutions*
10.4 Role of H2CO3 / HCO3- in controlling pH of blood
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In a healthy person, the pH of the blood is maintained at 7.35 to 7.45
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The main buffer in blood is H2CO3 / HCO3-
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When acid enters the blood,
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HCO3 -(aq) + H3O+(aq) → H2CO3(aq) + H2O(l)
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H2CO3(aq) is unstable and decomposes to form water and carbon dioxide
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H2CO3(aq) = CO2(g) + H2O(l)
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pH remains almost unchanged.
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When the base enters the blood,
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H2CO3(aq) + OH-(aq) → HCO3-(aq) + H2O(l)
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pH remains almost unchanged.
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10.5 Annex 1: Calculating pH of buffers
What is the pH of 50.00 mL buffer solution which is 2.00M in HC2H3O2 Iand 2.00M in NaC2H3O2?
Ans: Identify the acid-base pair, HC2H3O2 is the acid while NaC2H3O2 is its conjugate base.
pH = pKa+ log10([base]/[acid])
pH = -log10(Ka) + log10([2.00]/[2.00]) = 4.74 + 0.00 = 4.74
What is the new pH after 2.00 mL of 6.00M HCl is added to this buffer?
Ans: We have to understand that adding HCl means less conjugate base will be present while more of the weak acid will be formed.
Initial moles of acid and base in buffer is (2.00mol/L)(0.500L) = 0.100 mol
Initial moles of H+added is (6.00mol/L)(0.00200L) = 0.012 mol
Final moles of conjugate base = 0.1 - 0.012 = 0.088 mol
Final moles of weak acid = 0.012 + 0.1 = 0.112 mol
Final concentration of conjugate base = 1.69 M
Final concentration of weak acid = 2.15 M
pH = pKa+ log10([base]/[acid])
pH = -log10(Ka) + log10([1.69]/[2.15]) = 4.74 - 0.10 = 4.64
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