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77 Cards in this Set
- Front
- Back
Whenever a weak electrolyte and a strong electrolyte contain a common ion and are together in solution, the weak electrolyte ionizes less than it would if it were alone in solution |
This is the common ion effect |
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buffer solutions |
when small amounts of strong acid or strong base are added to solutions that contain a weak conjugate acid-base pair that resist big changes in pH when you add H+, OH- or water there is no single chemical that's a buffer |
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A buffer resists change in pH because |
it contains both an acid to neutralize OH- ions and a base to neutralize H+ ions but not enough to neutralize each other. |
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A buffer is prepared by |
mixing a weak acid or a weak base with a salt of that acid or base
ex: Buffer; CH3COOH-CH3COO- and adding CH3COONa to a solution of CH3COOH
or partially neutralizing a weak acid or a weak base (chemical reaction) mixing KHCrO4 and NaOH. |
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to calculate the pH of a buffer use |
pH=pKa + log[base]/[acid]
this is the Henderson-Hasselbalch equation
We can use the starting concentrations of the acid and base components of the buffer instead of the amounts that ionize. |
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Buffer capacity |
the amount of acid or base the buffer can neutralize before the pH changes a lot. It depends on the amount of acid and base used to prepare the buffer. the pH of a solution of weak electrolyte and its salt is the same if its concentration is different. ex: ph of 1M of CH3COOH and 1M CH3COONa is the same as the pH of 0.1M CH3COOH and 0.1M CH3COONa. However, the solution that has the greatest concentration has the greatest buffering capacity.
As long as we don't exceed the buffering capacity of the buffer, we can assume that the strong acid or strong base is completely consumed by reaction with the buffer. |
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the pH range of a buffer |
the pH range over which the buffer acts effectively
if the concentration of the weak acid or conjugate base is 10 times the conc. of the other than the buffering action is poor. because log10=1, buffers usually have a usable range within +\- pH unit of pKa, so pH=pKa +\- 1.
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It easier for a buffer to resist a change in pH when |
the concentration of a weak acid and its conjugate base are the same. |
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when the conc. of a weak acid and its conjugate base are equal |
pH=pKa. this gives the optimal pH of a buffer. |
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Explain why a mixture of acetic acid and sodium acetate can act as a buffer while a mixture of HCl and NaCl cannot. |
acetic acid and acetate are a weak conjugate acid/conjugate base pair which acts as a buffer because unionized acetic acid reacts with added base, while acetate combines with added acid leaving [H+] relatively unchanged. However Cl- wont combine with an acid to create HCl. any acid will increase the H+ conc. in an HCl and NaCl mixture. The conjugate bases of strong acids are negligible and mixtures of strong acids and their conjugate salts do not act as buffers. |
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what is acid-base titration? |
when a solution containing a known concentration of base is solwly added to an acid or visa versa. |
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pH titration curve |
a graph of the pH as a function of the volume of titrant added. used to determine the equivalence point and to select suitable indicators and to determine the Ka of the weak acid or the Kb of the weak base being titrated. |
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equivalence point of strong acid-base titration. |
an equal number of moles of base and acid have reacted leaving only a solution of their salt. The pH is 7 because the cation of a strong base and the anion of a strong acid are neither acids nor bases and therefore no effect on pH. |
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types of titrations |
1. strong acid strong base 2. weak acid strong base 3. polyprotic acid strong base |
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for a weak acid and strong base titration what is the initial pH and how is it found |
through Ka |
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difference between a titration curve for a weak acid strong ase titration from the curve for a strong acid strong base titration |
1. the solution of the weak acid has a higher initial pH than a solution of a strong acid of the same concentration. 2. The pH change in the rapid-rise portion of the curve near the equivalence point is smaller fro the weak acid than for the strong acid. 3. The pH at the equivalence point is above 7.00 for the weak acid titration. |
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end point |
the point in titration where the indicator changes color. is not the equivalence point. |
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pH change near the equivalence point becomes smaller as Ka decreases. |
... |
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saturated solution |
the solution is in contact with undissolved solute. |
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Ksp |
the solubility-product constant. it indicates how soluble the solid is in water. sp=solubility product. for heterogeneous equilibrium. its for the equilibrium between an ionic solid and its saturated solution and is a unitless number. its a measure of how much of the solid dissolves to form a saturated solution. has only one value for a given solute at any specific temperature
it is the solubility of an "insoluble" salt |
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solubility of a substance |
the quantity that dissolves to form a saturated solution. usually expressed as grams of solute per liter of solution g/L can change example; hydroxide salts are dependent upon the pH of the solution also solubility is affected by concentrations of other ions in solution: common ions. it changes as other species in solution change. |
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Molar solubility |
the number of moles of solute that dissolve in forming 1L of saturated solution of the solute mol/L |
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what effects solubility |
presence of common ions, solution pH, and presence of complexing agents. and amphoterism. |
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how is solubility affected by common ions |
the solubility of the salt decreases in the presence of common ions. however the value of Ksp is unchanged by the presence o additional solutes. |
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how does pH affect solubility |
it affects the solubility of any substance whose anion is basic. the affects are noticeable when one or both ions in the compound are at least moderately acidic or basic. The solubility of a compound containing a basic anion, the anion of a weak acid increases as the solution becomes more acidic. or as H+ increases/ pH is lowered. |
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A salt that has a conjugate base of a strong acid for exampe Cl-, the substances solubility is |
unaffected by changes in pH. |
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how formation of complex ions affect solubility |
lewis bases other than water can interact with metal ions particularly transition metal ions increasing their solubility, but the base must be able to interact more strongly with the metal ion than water does. ex of lewis base: NH3 |
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what is Q? |
the reaction quotient to determine the direction in which a reaction must proceed to reach equilibrium. is the mass action expression but with non equilibrium concentrations plugged in. Q is the molarity/concentrations of an "insoluble" salt or the the quotient for gases. |
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if Q>Ksp |
precipitation occurs, reducing ion concentrations until Q=Ksp reaction goes from right to left to reach equilibrium. because we know that Ksp is the dissociation equation for an insoluble ion so its products are the ions. |
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if Q=Ksp |
equilibrium exists, its a saturated solution |
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if Q |
solid dissolves increasing ion concentrations until Q=Ksp reaction proceeds from left to right to reach equilibrium. |
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selective precipitation |
separation of ions in an aqueous solution by using a reagent that forms a precipitate with one or more but not all of the ions. ex: Ag+ and Cu2+ in soln. If HCl is added to the soln. AgCl precipitates while Cu2+ remains in soln, because CuCl2 is soluble. |
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in selective precipitation, how do you know how much conc of reagent is needed for precipitation of each salt and which salt would precipitate first? |
the salt requiring the lower reagent ion conc precipitates first so you need to solve the reagent using Ksp.
In order to determine the amount needed you find the concentrations of both salts and and whatever it is, that is the range in which the reagent conc. is needed. |
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Kp |
equilibrium constant when dealing with partial pressures of gases in atmospheres. p=pressure. |
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know relationship between Kp and Kc |
Kp=Kc(RT)^Δn
delta n=the change in the number of moles of gas in the equation. =moles of gaseous product-moles of gaseous reactant. |
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know PV=NRT |
T=Kelvin n=moles p=usually in atm v=usually in liters R=0.08206 L-atm/mol-k |
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if K is greater than 1, large K |
equilibrium lies to right, products are predominate |
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if K is less than 1, small K |
equilibrium lies to left, reactants are predominate |
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Kp is proportional with Kc |
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what happens to Kc when the reaction is in the reverse direction of the original equation |
it becomes the inverse of the original Kc. |
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what happens to Kc when the reaction has been multiplied by a number |
the original equilibrium constant is raised to a power equal to that number. so if the reaction was multiplied by 2, Kc is raised to the second power. |
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when there is two or more equations that make up a net reaction what happens to Kc |
Kc becomes the product of the equilibrium constants for the individual reactions. |
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heterogeneous equilibria |
when substances in equilibrium are in different phases |
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Kc can be used if the substances are gases also, however if its given that they are in atm use Kp. |
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Complex ions |
usually transition metal ions react with lewis bases to make a complex ion. The lewis base must be able to interact more strongly with the metal ion than water and it increases the solubility of the metal salt. |
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Kf |
formation constant, used for complex ions where the product is the complex ion. |
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When Kf is large what does this mean? |
that the metal ion is assumed to have all converted to its complex ion, therefore the initial concentration of the metal ion would give the initial concentration of its complex ion. |
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Amphoteric oxides and amphoteric hydroxides |
some metal oxides and hydroxides that are insoluble in water dissolve in strongly acidic and strongly basic solutions because they behave as either an acid or base. examples: Al3+, Cr3+, Zn2+, Sn2+. They dissolve in acidic solutions because their anions react with acids and they can form complex anions. |
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Q< Kc |
reaction moves forward to produce more of the product because there are more reactants moves from left to right |
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Q>Kc |
reaction moves in reverse to produce more reactants because there are more products moves from right to left |
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For selective precipitation, precipitation starts when Q=Ksp so don't account for the molarities in the equation |
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Why is the concentration of undissolved solid not included in the Ksp expression? |
The concentration of undissolved solid doesn't appear in Ksp because solids already has a specific volume not dependent on the solution volume.
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what is the difference between solubility and Ksp |
Solubility is the amount grams or moles of solute that will dissolve in a certain volume of solution. Ksp is an equilibrium constant the product of the molar concentrations of all the dissolved ions in solution. |
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Metal hydroxides when pH becomes more basic their solubility usually |
decreases. |
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The greater the Ksp |
the greater the molar solubility if all is one to one in molarity otherwise solve by solving the molar solubility using Ksp. |
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Kc=Kf/Kr |
at equilibrium the rate of the forward reaction and reverse reaction are equal. |
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the more concentrated acid requires a greater volume of titrant to reach equivalence unless both acids have the same number of moles |
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the stronger the acid the greater the Ka and the larger its change in pH at the equivalence point. |
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difference between pH equivalence point of SA/SB vs. WA/SB |
SA/SB will have a pH of 7 because the salt is negligible it doesn't act as an acid or base, whereas in the weak acid its conjugate base acts as a base making the solution more basic and having a higher pH at the equivalence point. |
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After the equivalence point if the titrant is a base its OH- concentration at equilibrium determines pH. if its an acid the H+ determines the pH. Also after the equivalence point the number of moles of the conj acid or base is the same as at the equivalence point. |
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if pH=pKa |
conc of conj acid =conc of conj base |
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if pH |
the conc of weak acid is greater than conc of conj base. |
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if pH > pKa |
conc of conj base is greater than conc of weak acid. |
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to make a buffer with a certain pH you have to |
1. figure out what acid/base conjugate pair to use an ideal buffer has roughly equal concentrations of the weak acid and the weak base. when you are choosing a weak acid for your buffer, pick an acid where pKa of the acid is approximately equal to the pH you want in your buffer. 2. how much of each ingredient to use. Calculate the ratio between the acid and the base in the buffer.
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strong acid/base titration curve |
has a very steep change in pH at end point a large delta pH near end pt. a quite low initial pH pH is 7 at end pt. |
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there are two ways to write K for gases |
1. using pressures in atm PNH3 ex 2. you can also use moles/L |
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how do you write the equilibrium equation for a saturated solution |
the insoluble as the reactant and the ions as the products use Ksp. |
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to predict if a PPT will form use |
Q |
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complex ions can form when |
a metal cation bonds to ligands (lewis bases) |
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forming a complex ion does what to the concentration of the free metal ion |
it dramtically lowers it. |
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Buffer problems |
1. figure out what you're mixing and what's in solution. if its just a conj acid/base and its conj salt then its just a buffer in solution. look out for partial neutralization where its a weak acid or base that reacts with an H+ or OH-. 2. write an equilibrium equation 3. if something is added like HCl to a buffer or its a partial neutralization, write the equilibrium with mmoles to find the final Molarity. then afterwards write a Ka or Kb and write an equation for your buffer to find the pH. 4. x is usually small so you can always ignore it 5. you can tell if your answer is right by comparing the initial buffer's pH and the pH after something has been added to the buffer and seeing that the pH didn't change that much. |
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if its a buffer problem and you need to find volume given a pH and mL of the buffer. |
1. you know that the given pH will either give you the final conc at equilibrium of H+ or OH-. So write your Ka or Kb equation. then divide the Ka or Kb by the H+ or OH- to write a ratio for the buffer. 2. then given the mL of the solution this is your total volume. since you are given the initial Molarity of each substance, set an x indicating the amount of mL of either the acid or base doesn't matter which one and the other one do the total volume in ml -x. 3. solve for x and you found the volume of each in a buffer. |
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if a question asks if the reaction goes forward or backward given the number of moles of gases the equil problem, and the kC, how would you solve this problem and what type of problem is this? |
this is a gas problem where we have to compare Q with Kc. when it says forward or backward it means will it make more moles of products or more moles of the reactants. To solve it we first have to figure out the initial concentration of each gas because Q is the initial concentrations. Then we set up Q just like how we set up Kc. then we plug in the initial concentrations which were found without using a table and plug into the mass action expression. then compare it to Kc and figure out if Q>Kc which will favor the reverse reaction or if Q |
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what does it mean if Ksp is very low |
that there is a lot of PPT that forms and the concentrations of substances that do dissolve can be ignored if you have a solution with a common ion, use the concentration of that common ion instead. |
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when you see the word dissolve what should you think |
this is a Ksp problem. |
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How do you approach a Ksp problem |
1. understand what is in solution 2. know the difference between Q and Ksp, they are both set up the same way except Q isn't at equilibrium. 3. write the equilibrium equation 4. write the Ksp equation 5. set up a table if needed, we know that the insoluble solid will not create a final molarity given that it has its own fixed volume, so use only the moles/mmoles to solve the equation. |
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How can I increase the solubility of an insoluble compound ex: Ag2Co3 |
the only way to increase solubility is to lower the conc of either Ag+ or (Co3)2-
-to lower [Co32-] you have to add an acid which creates a second equilibria HCO3- > H+ + (Co3)2- adding H+ increase H+ conc and decreases the carbonate conc. therefore decrease the carbonate conc, increase the Ag+ conc and shifts the equilibrium to the right to increase the solubility of Ag2Co3
-To lower [Ag+], we need to create a silver complex ion because we know that a complex ion decrease the free ion conc.
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