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chemreviewexam3

# chemreviewexam3 - THE BEST REVIEW FOR EXAM 3 HOLCOMBE...

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THE BEST REVIEW FOR EXAM 3 -­૒ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER 1 Buffer: mixture of a weak acid and conjugate base or a weak base and conjugate acid. Buffers are solutions, which contain comparative amounts of a conjugate weak acid/base part and allow for only small changes in the pH of a chemical system, compared to non- buffered solutions, when an acid or a base is introduced. Buffer Capacity: amount of protons or hydroxide ions it can absorb without a significant change in pH. Mmol = 1 mole /1000 Henderson-ːHasselbalch equation: pH = pKa + log (A-­૒/HA) for a buffer – this is what you’ll use for the majority of problems that look like this… Ex1. What is the pH of a solution made to be 0.5 M in NH3 and 0.3 M in NH4Br? Kb for NH3 is 1.8 × 10 5. We first need to solve for Ka. Then we use the concentrations of NH3 and NH4+ in the equation. We always put the concentration of the molecule with more H ʼ s in it on the bottom – easiest way to remember how to do it. So pH = pKa + log (NH3/NH4). Simple as that. Ans: 9.48 Ex2. What is the pH of an aqueous buffer solution that is 0.1 M HF and 0.3 M KF? HF, pKa = 3.45 Same type of problem. More H ʼ s goes on bottom. pH= pka + log(KF/HF). Answer: 3.93

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THE BEST REVIEW FOR EXAM 3 -­૒ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER 2 Ex. Recognizing a buffer problem: What is the pH of an aqueous solution that is 0.10 M HCOOH (Ka = 1.8 × 10 4) and 0.10M NaHCO2? The key to this question is to realize that the solution contains a weak acid and its' conjugate base in equal concentrations, i.e., a buffer. For this type of solution, pH = pK a . Salts of form AB: We solve just like for a weak acid. Molar solubility is x = K sp 1/2 for salts of type AB Salts of form A 2 B or AB 2 : K sp = [A+][B-­૒] 2 = [x][2x] 2 = 4x 3 Molar solubility of type A2B and AB2 = (K sp /4) 1/3 Ex. Determine the molar solubility of calcium chloride (CaCl 2 ) if K sp = 2.5 × 10 2 . Salts of form AB 3 or A 3 B: 4 ions so take the 4 th root A saturated solution of MX4 (of MW 145 g/mol) contains 0.0175 grams per liter. What is the solubility product constant of MX4? I couldn ʼ t believe I missed this on the test. But I knew how to do it. I just plugged it in the wrong way. I know the Ksp = [x][4x] 4 but if you try to multiply the x ʼ s out and get 16X^5, then plug X in, you get the wrong answer. You have to plug in 4X and then put that to the 4 th power, then multiply that by x. x being the concentration you find from 0.0175/145. Don ʼ t miss this!!! K sp = Solubility Constant
THE BEST REVIEW FOR EXAM 3 -­૒ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER 3 When k is low or K is really small (10 -­૒28 ) everything “stays left”, which means the reaction does not dissociate and is insoluble. Ex. CaF 2 -­૒> Ca 2+ + 2F -­૒ K sp =10 -­૒12 so the reaction stays left, doesn’t diss. NaCl-­૒> Na+ + Cl-­૒ K = infinity so reaction goes to completion, to right Precipitations: Does a precipitate form when 100 mL of 0.00250 M AgNO3 and 100 mL of 0.00200 M NaBr solutions are mixed?

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