PH and pOH



Name:_____KEY_____________Chemistry: pH and pOH calculationsPart 1: Fill in the missing information in the table below. pH[ H3O1+ ]pOH[ OH1– ]ACID or BASE?3.781.65 x 10-4 M10.226.0 x 10-11 MACID3.413.89 x 10–4 M10.592.57 x10-11 MACID8.812.0 x 10-9 M5.196.46 x 10-6 MBASE8.692.04 x 10-9 M5.314.88 x 10–6 MBASE8.463.47 X 10-9 M5.542.88 x 10-6 MBASE12.078.45 x 10–13 M1.931.18 x 10-2 MBASE11.861.38 x 10-12 M2.147.2 x 10-3 MBASE3.364 .0 x 10-4 M10.642.31 x 10–11 MACID10.911.23 x 10-11 M3.098.0 x 10-4 MBASE5.137.49 x 10–6 M8.871.34 x 10-9 MACID4.068.6 x 10-5 M9.941.15 x 10-10 MACID6.413.89 x 10-7 M7.592.57 x 10–8 MACID4.166.92 x 10-5 M9.841.45 x 10-10 MACID0.971.06 x 10–1 M13.039.43 x 10-14 MACID10.186.6 x 10-11 M3.822.0 x 10-4 MBASE7.931.17 x 10-8 M6.078.53 x 10–7 MBASE7.058.9 x 10-8 M6.951.12 x 10-7 MBASE9.334.73 x 10–10 M4.672.11 x10-5 MBASE12.672.14 x 10-13 M1.334.68 x 10-2 MBASE12.001.00 x 10-12 M2.009.87 x 10–3 MBASE11.682.09 x 10-12 M2.324.8 x 10-3 MBASE7.049.22 x 10–8 M6.961.08 x 10-7 MBASE1.761.74 x 10-3 M12.245.75 x 10-13 MACID2.731.9 x 10-4 M11.275.39 x 10–12 MACIDpH = -log10 [H+] pOH = -log10 [OH-] pH + pOH = 14 [H+] = 10-pH [OH-] =10-pOH [H+][OH-] = 1.0 x 10-14Part 2: For each of the problems below, assume 100% dissociation.1. A.Write the equation for the dissociation of hydrochloric acid.HCl(aq) → H+(aq) + Cl-(aq)B.Find the pH of a 0.00476 M hydrochloric acid solution.pH = -log10 0.00476 pH = 2.322.A.Write the equation for the dissociation of sulfuric acid.H2SO4(aq) → H+(aq) + HSO4-(aq)HSO4-(aq) → H+(aq) + SO42-(aq)ORH2SO4(aq) → 2H+(aq) + SO42-(aq)B.Find the pH of a solution that contains 3.25 g of H2SO4 dissolved in 2.75 liters of solution.First, find the concentration: M = (mol/L)Mol = (3.25g/98g/mol) = 0.0332 molM = (0.0332mol/2.75L) = 0.0121M[H+] = 2( 0.0121) MpH = -log10 0.0242 = 1.623.A.Write the equation for the dissociation of sodium hydroxide.NaOH(aq) → Na+(aq) + OH-(aq)B.Find the pH of a 0.000841 M solution of sodium hydroxide.0.000841 M NaOH → [OH-] = 0.000841MpOH = -log10 [OH-] = -log10 0.000841 = 3.08pH = 14 – 3.08 = 10.924.A.Write the equation for the dissociation of aluminum hydroxide.Al(OH)3(aq) → Al3+(aq) + 3OH-(aq)B.If the pH is 9.85, what is the concentration of the aluminum hydroxide solution?pOH + pH =14pOH = (14- pH) = (14-9.85) = 4.15[OH-] = 10-pOH5.A.Write the equation for the dissociation of calcium hydroxide.Ca(OH)2(aq) →Ca2+(aq) + 2OH-(aq)B. If the pH is 11.64 and you have 2.55 L of solution, how many grams of calcium hydroxide are in the solution?pOH = (14-pH) = (14-11.64) = 2.36[OH-] = 10-pOH = 10-2.36 = 4.4 x 10-3MM = (mol/L)Mol NaOH = (M x L) = (4.4 x 10-3mol/L x 2.55L) = 0.0111Mass NaOH = (0.0111mol x 40g/mol) = 0.445 g ................
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