<p> CHEM 178 SI Wednesday, March 5th, 2008</p><p>1. What is the pH of a 0.20 M solution of hydrazinium chloride, N2H5Cl? Hydrazine, N2H4, is a weak -6 base with Kb = 1.7 x 10 . + + N2H5 + H2O  N2H4 + H3O Ka = Kw/Kb = 1.0 x 10-14 / 1.7 x 10-6 = 5.9 x 10-7 + Ka = [H+][ N2H4] /[ N2H5 ] Using an ICE table and assuming that x is very small will yield the relationship: + 2 Ka [ N2H5 ] = x x = (5.9 x 10-7 * 0.20)1/2 = 3.4 x 10-5 = [H+] pH = -log(x) = 4.46</p><p>2. A buffer solution contains 0.10 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L. Ka for acetic acid is 1.8 ´ 10−5 (a) What is the pH of this buffer? pH = pKa + log([base]/[acid]) pH = -log(1.8 ´ 10−5) + log(.13/.10) = 4.86</p><p>(b) What is the pH of the buffer after the addition of 0.02 M of KOH? (assume still in 1.0 L) - + - KOH CH3COOH CH3COO KOH  K + OH - - I 0.02 0.10 0.13 OH + CH3COOH  H2O + CH3COO C -.02 -.02 +.02 E 0 0.08 0.15 pH = pKa + log([base]/[acid]) pH = -log(1.8 ´ 10−5) + log(0.15/0.08) = 5.02</p><p>(c) What is the pH of the buffer after the addition of 0.02 M of HNO3? (assume still in 1.0 L) - + - HNO3 CH3COOH CH3COO HNO3  H + NO3 + - I 0.02 0.10 0.13 H + CH3COO  H2O + CH3COOH C -.02 +.02 -.02 E 0 0.12 0.11 pH = pKa + log([base]/[acid]) pH = -log(1.8 ´ 10−5) + log(0.11/0.12) = 4.71</p><p>3. What happens to the pH when NaCN is added to a solution of HCN? HCN  H+ + CN- NaCN  Na+ + CN- An increase in the concentration of a product will shift the reaction towards the reactant side, shifting this reaction away from producing H+, increasing the pH.</p><p>(b) If the initial concentration of HCN is 0.50M, what will the pH be after 0.40M NaCN is added to the solution? Ka = 4.9x10-10 pH = pKa + log([base]/[acid]) pH = -log(4.9x10-10) + log(0.40/0.50) = 9.21</p><p>4. Rank the following acids in order of increasing acid strength (weakest to strongest): H2SO4, H3PO4, H2SeO3, H2SO3, HClO4, HClO</p><p>HClO < H2SeO3 < H2SO3 < H3PO4 < H2SO4 < HClO4 </p>