CHM152GW5,6 S09 key

# Chemistry (MasteringChemistry Series)

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Unformatted text preview: CHM 152 Group Work 5 & 6 Spring 2009 Crib 1. Calculate the pH of a solution containing 0.285 moles of hypobromous acid (HBrO) and 0.315 moles of NaBrO in 2.50 liters of water. Given K a for HBrO is 7.8 x 10-7 . Show your work below. a. 6.06 b. 6.11 c. 6.15 d. 7.01 e. 0.545 What is in the beaker? A weak acid HBrO, and its conjugate base, BrO- ions from NaBrO. (Na + ions are spectators) So we have a buffer and can use the buffer equation. So we need the concentrations of these in the beaker. HBrO .285 moles / 2.50L = 0.114M HBrO (acid) NaBrO .315 moles / 2.50L = 0.126M NaBrO that dissociates 100% so = 0.126M BrO- (c. base) pH = 6.1079 + log( 0.126 / 0.114) = 6.15 (need 2 decimal places since Ka had two sig dig) 2. Give the formulas for two chemicals that would make a buffer solution in water. Like HF and KF 3. Calculate the pH when 15.0 mL of 0.150M perchloric acid is added to 12.0 mL of 0.125M potassium hydroxide. Show your work, balanced reaction and IF table for full credit. Strong acid and strong base. Reacts one way. HClO 4 (aq) + KOH (aq) J H 2 O (l) + KClO 4 (aq) Need moles of each. acid: 0.0150L ( 0.150 mol / L ) = 0.00225 moles acid base: 0.0120L (0.125mol / L) = 0.00150 moles base Set up initial final table HClO 4 + KOH J H 2 O (l) + KClO 4 I 0.00225 moles 0.00150 moles --- 0 - 0.00150 all reacts --- + 0.00150 F 0.00075 moles 0 --- 0.00150 moles NOT a buffer by the way!!! KClO 4 is a neutral salt, not a conjugate base. Note the new volume is 27.0 mL. pH will depend on the strong acid left over not the neutral salt. [H + ] = 0.00075 moles / 0.0270L = 0.0278M pH = 1.56 (final answer needs 2 decimal places since 0.00075 moles had two sig dig) 4. How many milliliters of 0.35M lithium hydroxide must be added to 25.0 mL of 0.45M acetic acid to reach the equivalence point? What is the pH at the equivalence point? Given: K a for acetic acid is 1.8 x 10-5 Weak acid and strong base. We are interested in the equivalence point. Stoichiometry!!! 25.0 mL acetic ( 0.45 moles / 1000mL)(1 LiOH / 1 acetic)(1000mL / 0.35moles LiOH) = V b V b = 32.143 mL = 32.1 mL First they react together one way since NaOH is strong. Set up an initial final table. Calculate moles of each. Note they are equal since we are at the equivalence point....
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CHM152GW5,6 S09 key - CHM 152 Group Work 5 & 6 Spring...

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