CHEMISTRY CHAPTER 16 – Flashcards
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weak acids or conjugate bases that can react with strong acids or bases to prevent sharp, sudden changes in pH
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buffers
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Ethylene Glycol, A chemical that is mixed with water and added to the cooling system of a car. It moves through the engine to remove extra heat.
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antifreeze
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a compound with the formula H2CO3 that results from the combination of carbon dioxide (CO2) and water (H2O); of particular importance in maintaining the body's acid-base balance.
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carbonic acid
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abnormally high acidity (excess hydrogen-ion concentration) of the blood and other body tissues
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acidosis
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an ion containing protein that transports the bulk of oxygen that is carried in the blood
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hemaglobin
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To provide ventilations at a higher rate than normal.
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hyperventilate
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what is happenining here
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carbonic acid neutralize base
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a weak acid by itself isnt strong enough to be a buffer and vice versa for weak bases
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true
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what must a buffer contain significant amounts of
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a weak acid and its conjagate base
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what would occur to the solution
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it will neutralize only if the strong base is amount is less
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The phenomenon in which the addition of an ion common to two solutes brings about precipitation or reduced ionization, Ex: AgCl in water If you add in NaCl (more Cl ion) then LESS solid will dissolve at equilbirum Application of Le Chatelier DECREASES solubility
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common ion effect
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write the three equations
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an equation that allows use to quickly find the ph of a buffer from the initial morality only if x is small
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H H equation
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pH = pKa + log ([A-]/[HA]) [A-] concentration of conjugate base [HA] concentration of acid Ka = dissociation constant
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henderson haselbach equation
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-logka, acid dissociation constant
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pka
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2) What is a buffer? How does a buffer work? How does it neutralize added acid? Added base?
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A buffer is a solution that is resistant to changes in pH. Buffers work by converting strong acids or bases into weak acids or bases. Consider a buffer that contains a weak acid (HA) and its conjugate base (A-). If we add a strong acid like HCl, or a strong base like NaOH, the reactions that occur are as follows HCl(aq) + A-(aq) ® HA(aq) + Cl-(aq) NaOH(aq) + HA ® H2O(l) + A-(aq)
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4) What is the Henderson-Hasselbalch equation and why is it useful?
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4) The Henderson-Hasselbach equation is pH = pKa + log10{[base]/[acid]} It is useful in preparing a buffer solution with a particular value for pH.
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27) In which of these solutions will HNO2 ionize less than it does in pure water? a) 0.10 M NaCl b) 0.10 M NaOH c) 0.10 M KNO3 d) 0.10 M NaNO2
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1) Which of the following reactions will go essentially to completion? a) Reaction of a strong acid with a strong base b) Reaction of a strong acid with a weak base c) Reaction of a weak acid with a weak base d) Both a and b e) Both a and b and c
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d
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2) Which of the following aqueous solutions would be a buffer solution? a) A solution that has 0.100 M HNO2 (a weak acid) and 0.100 M NO2- b) A solution that has 0.100 M HClO4 (a strong acid) and 0.100 M ClO4- c) A solution that has 0.100 M NH3 (a weak base) and 0.100 M NH4+ d) Both a and b e) Both a and c
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e
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3) Consider the titration of H3PO4, a weak polyprotic acid, with Ba(OH)2, a strong base. The number of equivalence points that will be observed in this titration is a) 3 b) 2 c) 1 d) 0 e) None of the above
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a
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1) The following questions refer to the titration curve given below. For each of the following questions circle the correct answer. There is one and only one correct answer per question. [3 points each] a) The titration curve shows the titration of a strong acid a weak acid a strong base a weak base with a strong base with a strong base with a strong acid with a strong acid b) Which point on the titration curve represents a region where a buffer solution has formed? point A point B point C point D c) Which point on the titration curve represents the equivalence point? point A point B point C point D d) Which of the following would be the best indicator to use in the titration? erythrosin b methyl red bromthymol blue o-cresonphthalein pKa = 2.9 pKa = 5.4 pKa = 6.8 pKa = 9.0
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a. weak acid, strong base b.point B c.point C d.o-cresonphthalein pKa = 9.0
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2) Define the following terms [4 points each] a) endpoint
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2) Define the following terms [4 points each] a) endpoint The point in a titration where the indicator being used changes color. One tries to choose an indicator whose end point is as close as possible to the equivalence point of the titration.
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2) Define the following terms [4 points each] b) selective precipitation
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b) selective precipitation The process where a substance is added to a solution that causes one solute in the solution to form a solid (precipitate) while the remaining solutes remain in solution. The precipitate can then be removed by filtration. Selective precipitation therefore is a method that can separate a particular solute from a mixture of solutes in a solution.
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3) A buffer solution is formed using formic acid (HCOOH, Ka = 1.8 x 10-4) and formate ion (HCOO-). The concentration of formic acid in the buffer solution is twice the concentration of formate ion. What is the pH of the buffer solution? [4 points]
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For a buffer, pH = pKa + log10{[base]/[acid]} But [acid] = 2 [base], and so pH = - log10(1.8 x 10-4) + log10 (1/2) = 3.44
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1) A chemist has a 0.2000 M stock solution of hypochlorous acid (HClO, a weak acid, with Ka = 2.9 x 10-8) and solid potassium hypochlorite (KClO, M = 90.55 g/mol), a strong electrolyte containing the conjugate base of hypochlorous acid. How many grams of KClO must be added to 1.000 L of the stock HClO solution to form a buffer solution with pH = 7.40? You may assume that the volume of the solution does not change when KClO is added. [12 points]
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pH = pKa + log10{[base]/[acid]} and so log10{[base]/[acid]}= pH - pKa = 7.40 - {- log10(2.9 x 10-8)} = - 0.138 and so (taking the inverse log10 of both sides of this equation) {[base]/[acid]} = 10-0.138 = 0.728 [base] = 0.728 [acid] = 0.728 (0.2000 M) = 0.1457 M Since potassium hypochlorite is a strong electrolyte (completely ionizes in water), then moles KClO = 1.000 L 0.1457 mol = 0.1457 mol 1.000 L grams KClO = 0.1457 mol 90.55 g = 13.2 g 1.00 mol ,
