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198 chemistry nicotine all behave as very weak bases due to K = 10 –4.75 = 1.77 × 10 M –5 their very small K b. Ammonia produces OH in – b NH + H O NH + + OH – aqueous solution: 3 2 4 + – Initial concentration (M) NH 3(aq) + H O(l2 NH (4q) + OH (aq) 0.10 0.20 0 The pH scale for the hydrogen ion concentration has been extended to get: Change to reach pK = –log (K ) (6.34) equilibrium (M) b b –x +x +x At equilibrium (M) Problem 6.21 0.10 – x 0.20 + x x The pH of 0.004M hydrazine solution is 9.7. + – Calculate its ionization constant K and pK . K b [NH ][OH4] / [NH ] 3 b b –5 = (0.20 + x)(x) / (0.1 – x) = 1.77 × 10 Solution AsK issmall,wecanneglectxincomparison b NH NH + H O NH NH + + OH – to 0.1M and 0.2M. Thus, 2 2 2 2 3 From the pH we can calculate the hydrogen [OH ] = x = 0.88 × 10 –5 + –9 ion concentration. Knowing hydrogen ion Therefore, [H ] = 1.12 × 10 concentration and the ionic product of + pH = – log[H ] = 8.95. water we can calculate the concentration of hydroxyl ions. Thus we have: 6.11.5 R elation between K a and K b [H ] = antilog (–pH) As seen earlier in this chapter, K and K a b = antilog (–9.7) = 1.67 ×10 –10 represent the strength of an acid and a base, [OH ] = K / [H ] = 1 × 10 –14 / 1.67 × 10 –10 respectively. In case of a conjugate acid-base w –5 pair, they are related in a simple manner so = 5.98 × 10 that if one is known, the other can be deduced. The concentration of the corresponding Considering the example of NH + and NH hydrazinium ion is also the same as that 4 3 of hydroxyl ion. The concentration of both we see, + + these ions is very small so the concentration NH 4(aq) + H O(l2 H 3 (aq) + NH (aq) 3 of the undissociated base can be taken + + –10 K a = [H 3 ][ NH ] / 3NH ] = 5.64× 10 equal to 0.004M. + – NH (3q) + H O(l)2 NH (4q) + OH (aq) Thus, + – –5 + – K b =[ NH ][4OH ] / NH = 1.8 × 30 K b [NH NH ]2OH ] 3 [NH NH ] 2 2 Net: 2 H O(l) H O (aq) + OH (aq) – –5 2 –7 2 3 = (5.98 × 10 ) / 0.004 = 8.96 × 10 K = [H O ][ OH ] = 1.0 × 10 –14 M –7 w 3 pK =b–logK = –lob(8.96 × 10 ) = 6.04. Where, K represents the strength of NH + a 4 Problem 6.22 as an acid and K b represents the strength of NH as a base. Calculate the pH of the solution in which 3 0.2M NH 4Cl and 0.1M NH are pre3ent. The It can be seen from the net reaction that pK of ammonia solution is 4.75. the equilibrium constant is equal to the b Solution product of equilibrium constants K a and K b for the reactions added. Thus, NH + H O NH + + OH – K × K = {[H O ][ NH ] / [NH ]} × {[NH ] + 3 2 4 a b 3 3 4 4 The ionization constant of NH , [OH ] / [NH ]} 3 3 K b = antilog (–pK ) i.eb = [H O ][OH ] = K 3 w = (5.6 ×10 –10) × (1.8 × 10 ) = 1.0 × 10 –14 M Reprint 2026-27