CHM152GW4 Su09 crib

Chemistry (MasteringChemistry Series)

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CHM 152 Group Work 5 (over Ch. 15) Crib 1. Calculate the pH of a 2.00 L solution containing 0.885 moles of hypochlorous acid (HClO) and 0.905 moles of NaClO. Given K a for HClO is 3.0 x 10 -8 . What is in the beaker? A weak acid HClO, and its conjugate base, ClO - ions from NaClO. (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. HClO .885 moles / 2.00L = 0.4425M HClO (acid) NaClO .905 moles / 2.00L = 0.4525M NaClO that dissociates 100% so = 0.4525M ClO - (c. base) pH = 7.5229 + log( 0.4525 / 0.4425) = 7.53 (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 25.0 mL of 0.100M HBr is added to 15.0 mL of 0.100M LiOH. Strong acid and strong base. Reacts one way. HBr (aq) + LiOH (aq) J H 2 O (l) + LiBr (aq) Need moles of each. acid: 0.0250L ( 0.100 mol / L ) = 0.00250 moles acid base: 0.0150L (0.100mol / L) = 0.00150 moles base Set up initial final table HBr + LiOH J H 2 O (l) + LiBr 0.00250 moles 0.00150 moles --- 0 - 0.00150, not all gone = excess all reacts, limiting - 0.00150 --- + 0.00150 .00100 mol 0 --- 0.00150 moles NOT a buffer by the way!!! LiBr is a neutral salt, not a conjugate base. Note the new volume is 40.0 mL. pH will depend on the strong acid left over not the neutral salt. So HBr dissociates 100%. Thus [H + ] = 0.00100 moles / 0.0400L = 0.0250M pH = 1.602 (final answer needs 3 decimal places since everything had three sig dig) 4. How many milliliters of 0.95M sodium hydroxide must be added to 35.0 mL of 0.85M acetic acid to reach the equivalence point? What is the pH at the equivalence point?
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CHM152GW4 Su09 crib - CHM 152 Group Work 5(over Ch 15 Crib...

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