CHM152GW4 S09 key

Chemistry (MasteringChemistry Series)

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Group Work 4, CHM 152, Key 1. Draw what the following chemicals would look like when added to a beaker of water. nitric acid acetic acid barium hydroxide ammonia 2. Now write the chemical formulas for the 4 chemicals above. HNO 3 CH 3 COOH Ba(OH) 2 NH 3 3. Now write the balanced chemical equation for what each of them do when put in water. nitric acid’s rxn: __HNO 3 (aq) + H 2 O(l) J NO 3 - (aq) + H 3 O + (aq) acetic acid’s rxn: __ CH 3 COOH (aq) + H 2 O(l) ' CH 3 COO - (aq) + H 3 O + (aq) barium hydroxide’s rxn: __ Ba(OH) 2 (s) J Ba +2 (aq) + 2 OH - (aq) ammonia’s rxn: __ NH 3 (aq) + H 2 O(l) ' NH 4 + (aq) + OH - (aq) 4. If the pH is 11.248, calculate the pOH, [H + ], and [OH - ] values. pOH = _ 2.752 [H + ] = _ 5.65 x 10 -12 M [OH - ] = _ 1.77 x 10 -3 M_ 5. Calculate the pH, pOH, [H
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Unformatted text preview: + ], and [OH-] values for a 0.0525M solution of hydroioidic acid. HI is strong and ionizes about 100% so at equilibrium [H+] = 0.0525M so pH = 1.280 then pOH = 12.720 and [OH-] = 1.90 x 10-13 M 6. Calculate the pH for a 1.35 M solution of a weak acid HA with K a = 8.39 x 10-5 . This is a weak acid so equilibrium is reached. HA(aq) + H 2 O(l) ' A -(aq) + H 3 O + (aq) I 1.35 - 0 0 C -x - +x +x E (1.35-x) - (x) (x) K a = 8.39 x 10-5 = x 2 / (1.35-x) approx = x 2 / 1.35 8.39 x 10-5 (1.35) = x 2 take the square root of both sides x = 0.01064 which is also H+ so pH = -log x = 1.973 = pH check approximation 0.01064 / 1.35 x 100% = 0.788% which is less than 5% so valid...
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This document was uploaded on 09/27/2011.

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