00500x x2 ka x2 000500 18 104x 36 105 0 x2

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Unformatted text preview: 3 0.00500 0.200 [Formic H] –3 x = 7.2 × 10 = 3.6% < 5% OK. –3 x 0.2 [Formic acid]in = 7.2 × 10 = 144%!! NOT OK [Formate]in 0.005 14 Can’t make that assumption. Go back and recalculate x Initial composition Change in composition Equilibrium composition formic acid 0.200 –x 0.200 – x formate ion 0.00500 +x 0.00500 + x H3O+ 0 +x x [H3O+][Formate–] (0.00500 + x)x 0.00500x + x2 = = = 1.8 × 10–4 0.200 – x 0.200 – x [Formic H] 3.6 × 10–5 – 1.8 × 10–4 x = 0.00500x + x2 Ka = x2 + (0.00500 + 1.8 × 10–4)x – 3.6 × 10–5 = 0 x2 + 0.00518x – 3.6 × 10–5 = 0 – 0.00518 ± √(0.00518)2 – 4(1)(–3.6 × 10–5) x= = 3.94515 × 10–3 or –0.0091251 2(1) [H3O+] = x = 3.94515 × 10–3 pH = –log [H3O+] = –log(3.94515 × 10–3) = 2.404 15 Friday, March 1, 13 What happens when we add acid or base to a buffer solution? Reacts Reacts Reduces Increases with acid with base Weak acid/conjugate base buffer: HA(aq) + H2O(aq) ← H3O+(aq) + A–(aq) → A–(aq) + H2O(aq) ← OH–(aq) + HA(aq) → [HA] [A–] √ [A–] [HA] √ [B] [BH+] [BH+] [B] √ Weak base/conjugate acid buffer: B(aq) + H2O(aq) ← OH–(aq) + BH+(aq) → BH+(aq) + H2O(aq) ← H3 → O+(aq) + B(aq) √ 16 To perform calculations on acids or bases added to buffers: 1. Stoichiometric calculations to determine the concentrations of components after the added base/acid is neutralized (see reduces/ increases columns) 2. Use HH law, unless a large amount of acid/base is added, or if the equilibrium constant of one of the reactions in the buffer is too high (meaning it isn’t a very weak acid or base), or the amount of one of the components isn’t very large, in which case you’ll need to perform an equilibrium calculation. 17 Calculation Time A solution has an initial concentration of 0.0100 M HClO (Ka = 3.5 × 10–8) and 0.0300 M NaClO. What is the pH after the addition of 0.0030 mol of solid NaOH to 1.00 L of this solution, assuming no volume change. The two equilibria in the buffer are: HClO(aq) + H2O(l) ← H3O+(aq) + ClO–(aq) → ClO–(aq) + H2O(l)...
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