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gen chem ch 16
| Question | Answer |
|---|---|
| buffers are solutions that | resist changes in pH when an acid or base is added. They can switch back and forth. If the solution gets too basic, the buffer takes an acid approach. It maintains pH. |
| They act by neutralizing the | added acid or base |
| But just like everything else, there is a limit to what buffers can do | eventually the pH changes. |
| A buffer contains significant amounts of | both a weak acid and conjugate base (or a weak base and its conjugate acid.) |
| Many buffers are made by mixing a solution of | a weak acid with a solution of soluble salt containing its conjugate base anion. |
| The weak acid neutralizes | added base. |
| The conjugate base neutralizes | added acid. |
| Buffers work by applying | Le Châtelier’s Principle to weak acid equilibrium. |
| A buffer must contain both what and what to be a buffer? | weak acid and weak base. |
| Buffer solutions contain significant amounts of the __________, ____- these molecules react with added ______ to neutralize it. | weak acid molecules, HA – these molecules react with added base to neutralize it. |
| You can also think of the H3O+ combining with the OH− to make H2O; the H3O+ is then replaced by the | shifting equilibrium |
| HA(aq) + H2O(l) <--> A−(aq) + H3O+(aq) The buffer solutions also contain significant amounts of the conjugate base anion, A− - these ions combine with added acid to make more | HA and keep the H3O+ constant. |
| The Henderson-Hasselbalch Equation | calculating the pH of a buffer solution can be simplified by using an equation derived from the Ka expression called the Henderson-Hasselbalch Equation. |
| The Henderson-Hasselbalch Equation calculates | the pH of a buffer from the Ka and initial concentrations of the weak acid and salt of the conjugate base. As long as the “x is small” approximation is valid |
| How can you tell which one is the acid and which one is the base? | the acid will have one more H+ than the base. |
| What is the equation for getting pH? | pH=-log[H3O+] |
| What does pKa equal? | pKa=-logKa. |
| The Henderson-Hasselbalch equation is generally good enough when the “x is small” approximation is applicable. Generally, the “x is small” approximation will work when both of the following are true | 1) the initial concentrations of acid and salt are not very dilute. 2) the Ka is fairly small. For most problems, this means that the initial acid and salt concentrations should be over 1000x larger than the value of Ka |
| Though buffers do resist change in pH when acid or base are added to them, their pH does | change. |
| Calculating the new pH after adding acid or base requires breaking the problem into 2 parts. 1) | 1. a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other, added acid reacts with the A− to make more HA. Added base reac |
| If an WEAK acid is added to the buffer, what will happen? | to neutralize the acid, it will take the same amount of base. Therefore, you will add however much acid is added, and subtract that amount from the base. Then you add that amount to the HA(aq) on the other side of the equation. |
| Calculating the new pH after adding acid or base requires | breaking the problem into 2 parts. 2) an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]. |
| What is the equation when adding a weak acid to a buffer? | 1) The adding table= H+(aq)+A-(aq)--> HA(aq) 2) The RICE Table= HA+H20 <--> H3O+A- |
| What is the equation when adding a weak base to a buffer? | 1)The adding table= OH-+HA --> H2O+A-. Ex= OH 2 +HC2H3O2--> H2O+ C2H3O2 - 2) The RICE Table= HC2H3O2+H2O<--> H3O + +C2H3O2 |
| When you calculate the pH of a buffer after adding small amounts of acid or base, remember; Adding a small amount of strong acid to a buffer converts a stoichiometric amount of the base to the | conjugate acid. |
| When you calculate the pH of a buffer after adding small amounts of acid or base, remember; Adding a small amount of strong base to a buffer converts a stoichiometric amount of the acid to | the conjugate base. |
| Stoichiometric calculation | the table used is not a ICE table. This table simply tracks the stoichiometric changes that occur during the neutralization of the added acid. |
| pKa + pKb= | 14 |
| Equivalence point | the point in a titration at which the added solute completely reacts with the solute present in the solution. *This is the point where the amount of acid is stoichimetrically equal to the amount of base in solution. |
| End point | the point of pH change where an indicator changes color. |
| Ksp is | solubility product constant. (it will be given). |
| In the equation AgCl <--> Ag+ + Cl- what is the Ksp expression? | [Ag+][Cl-] |
| Capacity of a buffer | how much added acid or base it can effectively neutralize. |
| Range of the buffer | the pH range over which a particular acid and its conjugate base can be effective. |
| A buffer is most effective or most resistant to pH changes when the concentrations of | acid and conjugate base are equal. |
| In order for a buffer to be reasonably effective, the relative concentrations of acid and conjugate base should not differ by more than a factor of | 10. |
| A buffer is most effective or most resistant to pH changes when the concentrations of acid and conjugate base are | high. |
| How do you find the percent change of pH? | you do Henderson’s Hasselbalch equation. When you get that answer, you subtract the initial concentration of the solution by it and then divide it all by the initial concentration. Then multiply by 100%. |
| The buffer with greater amounts of acid and conjugate base is more resistant to pH changes and therefore the more | effective buffer. |
| The more dilute the components, the | less effective the buffer. |
| To calculate the range of a pH. The LOWEST pH for effective buffer occurs when the base is | 0.10 or one tenth as concentrated as the acid. Use the Henderson Hasselbalch equation- pH=pKa + log (0.10). =pKa-1 |
| To calculate the range of pH. The HIGHEST pH for effective buffer occurs when the | base is 10 times as concentrated as the acid. pH= pKa + log10. =pKa+1 |
| The effective range for a buffering system is | one pH unit on either side of pKa. |
| Buffer capacity is the amount of | acid or base that you can add to a buffer without causing a large change in pH. |
| What does buffer capacity mean? | the ability to neutralize added acid and added base. |
| The buffer capacity increases with increasing | absolute concentrations of the buffer components. The more concentrated the weak acid and conjugate base that compose the buffer, the higher the buffer capacity. |
| Also, the overall buffer capacity increases as the relative concentrations of the | buffer components become more similar to each other. As the ratio of the buffer components gets closer to 1, the overall capacity of the buffer, the ability to neutralize added acid and added base, becomes GREATER. |
| When is the equivalence point with a STRONG ACID AND A STRONG BASE reached? | whenever the number of moles of base added equals the number of moles of acid initially in solution. |
| 1 mol of NaOH neutralized 1 mol of | H3O+. |
| The unit of H3O+ is | M. Which is mol/L. |
| We can calculate the H3O+ concentration by dividing the number of moles of H30+ remaining by the | total volume. The total volume is (the initial volume in Liters plus the added volume in L. This would make it M (or moles/L). |
| The pH at the equivalence point of a strong acid-strong base titration will always be | 7.00 (at 25 degrees Celsius). |
| We calculate the OH- concentration by | dividing the number of moles of OH- remaining by the TOTAL VOLUME (which it initial volume plus added volume). |
| The overall pH curve for a strong base and a strong acid is | it points left to right. The change between the base and acid is tall and quick. |
| Summarizing the titration of a strong acid with a strong base- the initial pH is simply the pH of the | strong acid solution to be titrated. |
| Summarizing the titration of a strong acid with a strong base- before the equivalence point, H3O+ is in excess. Calculate the [H3O+] by | subtracting the number of moles of added OH- from the initial number of moles of H30+ and dividing by the total volume. |
| Summarizing the titration of a strong acid with a strong base- at the equivalence point, neither reactant is in excess and the pH= | 7.00. |
| Summarizing the titration of a strong acid with a strong base- Beyond the equivalence point, OH- is in excess. Calculate the [OH-] by | subtracting the initial number of moles of H3O+ from the moles of added OH- and dividing by the total volume (initial plus final volume). |
| A buffer will be most effective when the [base] and [acid] are | 1) Equal concentrations of acid and base. 2) Effective when 0.1 < [base] and [acid] < 10. 3) A buffer will be most effective when the [acid] and the [base] are large. |
| For any buffer in which the amounts of weak acid and conjugate base are equal, the pH= | pKa. |
| The titration of a weak acid by a strong base will always have a ________ equivalence point | basic equivalence point because at the equivalence point, all of the acid had been converted into its conjugate base, resulting in a weakly basic solution. |
| The overall pH curve for the titration of a weak acid with a strong base has a | characteristic “S-shaped” similar to that for the titration of a strong acid with a strong base. It goes down left and then up right. Remember that at the equivalence point the pH is more basic. That is why it is higher up on the pH chart on pg 737. |
| The initial pH is that of the weak acid solution. Calculate like a | weak acid equilibrium problem e.g., 15.5 and 15.6. |
| Between the initial pH and the equivalence point, the solution becomes a buffer. Use the reaction stoichiomiometry to calculate the amounts of each buffer component and then | use the Henderson-Hasselbalch equation to calculate the pH (as in Example 16.3)before the equivalence point, the solution becomes a buffer. |
| Halfway to the equivalence point, the buffer components are | exactly equal and pH=pKa. |
| At the equivalence point, the acid has all been converted into its conjugate base. Calculate the pH by working an equilibrium problem for the ionization by | the ion acting as a weak base (similar to Ex 15.14) (Calculate the concentration of the ion acting as a weak base by dividing the number of moles of the ion by the total volume at the equivalence point.) |
| Beyond the equivalence point, OH- is in excess. | Ignore the weak base and calculate the OH- by subtracting the initial number of moles of H3O+ from the number of moles of added OH- by dividing by the TOTAL VOLUME. |
| half-neutralization pH = | pKa |
| beyond equivalence point, the OH is in excess | [OH−] = mol MOH xs/total liters. [H3O+][OH−]=1 x 10-14 |
| Summarizing titration of a weak acid with a strong base- the initial pH is that of the weak acid solution to be titrated. Calculate the pH by working an equilibrium problem | (similar to Ex 15.5 and 15.6.) using the concentration of the weak acid as the initial concentration. |
| Summarizing titration of a weak acid with a strong base- between the initial pH and the equivalence point, the solution becomes a buffer. Use the reaction stoichiometry to calculate | the amount of each buffer component and then use the Henderson-Hasselbalch equation to calculate the pH (as in Example 16.3) . |
| Summarizing titration of a weak acid with a strong base- Halfway to the equivalence point, the buffer components are | exactly equal and the pH=pKa. |
| Summarizing titration of a weak acid with a strong base- At the equivalence point, the acid has all been converted into its conjugate base. Calculate the pH | by working an equilibrium problem for the ionization of water by the ion acting as a weak base. Calculate the concentration of the ion acting as a weak base by dividing the number of moles of the ion by the total volume at the equivalence point. |
| Summarizing titration of a weak acid with a strong base- Beyond the equivalence point, OH- is in excess. Ignore the weak base and calculate the OH- by | subtracting the initial number of moles of H3O from the number of moles of added OH- and dividing by the total volume. |
| Steps in how to do a weak acid-strong base titration pH curve. step 1 | (pg 738)1) the equivalence point occurs when the amount (in MOLES) of added base=the amount of(in MOLES) of acid initially in the solution. |
| Steps in how to do a weak acid-strong base titration pH curve. step 2 | 2) Begin by calculating the amount (in MOLES) of acid initially in the solution. The amount (in MOLES) of KOH that must be added to equal to the amount of the weak acid. |
| Steps in how to do a weak acid-strong base titration pH curve. step 3 | 3) Calculate the volume of KOH required from the number of moles of KOH required from the number of moles of KOH and the molarity. |
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