*High Yield GChem: Acids and Bases Flashcards

Question

What is acid-base titration? | What would one not use for titration?

Answer 1

Experimental technique used to determine the identity of an unknown weak acid (or weak base) by determining its pKa (or pKb). Also used to determine concentration of any acid or base solution Add titrant (strong acid or base) or known identity and concentration to solution of unknown base or acid (one never titrates a acid with another acid/ base with base) titrant is slowly added

Answer 2

increase Buffer region is plato, pH changes gradually bc HF was getting neutralized Equivalence point = drastic pH increase at which moles OH- added = moles H+ initially present, when equal concentrations of titrant and sample -> [H+] = [OH-] in this ex. just enough NaOH was added to completely neutralize all HF (all HF reacted with NaOH, so we just have F- and Na+) This solution is basic bc Marky said that at this point since HF is weak acid, all of it has reacted and it's conjugate base is left over so basic solution For weak acid titrated with strong base -> the equivalence point will occur pH > 7 For weak base titrated with strong acid -> the equivalence point will occur pH < 7 For strong acid titrated with strong base or strong base titrated with strong acid pH = 7 The pH at the equivalence point can tell us weather the acid or base we are titrating is weak or strong Half-equivalence point is middle of flat buffer region, it is half as much as titrant It represents when in this ex [HF] = [F-] and henderson-hasselbach tells us the pH = pKa of HF

Answer 3

the number of ionizable hydrogens the acid can donate | Weak base curve follows same general pattern expect its flipped bc at volume of 0 it starts at a higher pH

Answer 4

a * [A] * Va = b * [B] * Vb a -> number of acidic hydrogens per formula unit b -> how many H+ ions the base can accept The "MaV formula"

Answer 5

``` 5.3 Find Ka Make ICE table solve for x then plug into pH equation and rearrange ```

Answer 6

average of 2 half equivalence points (which are each at buffer plato part of graph bc they rep pKa of OG substance there, thing being added to) It is the zwitter ion, when aa is neutral, pH = pI

Answer 7

An indicator is a weak acid that is one color in its protonated form and other in its deprotonated form, since it is a weak acid it as a pKa associated with it and when the pKa is equal to the pH of the solution it is in, it will start to deprotonate, so you select an indicator with a pKa that's close to your target pH and see when you have added enough of your titrant when you observe a color change, so using indicators allows us to monitor titrations w/o machines and see exact moment when color change occurs, some indicators are better than others, ex the brom. blue has pKln = 7 equivalence point is moment when color change occurs *Within 1 unit of the pKa of the indicator, the end point represents the point of color change of the indicator end point The point in the titration process which is indicated by color change of the indicator is called endpoint.

Answer 8

Can see ipad goodnotes for work or ask Marky

Answer 9

B. A 2 M solution of the potent organic pollutant sodium methylthiolate (NaSCH3) is titrated to its endpoint with 2 M HCl. –log (1 × 10–5.2) best approximates the pH of the resultant solution. (pKb of SCH3– = 3.6) The result of the titration will be a 1 M solution of methylthiol, HSCH3 since the initial 2 M solution has been doubled in volume during the titration. Since we know the pKb of its conjugate base SCH3– is 3.6, then we know the pKa of HSCH3 is 10.4. Therefore, at equilibrium: 10–10.4 = (x)(x)/(1–x) Where x = [H+] = [SCH3–], at equilibrium. As the resultant value of x is negligibly small compared to 1, we ignore x in the denominator and say that x = [H+] = [SCH3–] = (10–10.4)(0.5) giving 1 × 10–5.2. The answer –log (2 × 10–5.2) would be correct if the resultant solution was 2 M in concentration. The answer –log (1 × 10–1.8) is the result of using the pKb of the methylthiolate anion in the calculation rather than the pKa of HSCH3, and –log (1 × 10–3.6) would be the result of a similar a calculation using this pKb value, while not taking the square root.

Answer 10

B. A 2 M solution of the potent organic pollutant sodium methylthiolate (NaSCH3) is titrated to its endpoint with 2 M HCl. –log (1 × 10–5.2) best approximates the pH of the resultant solution. (pKb of SCH3– = 3.6) The result of the titration will be a 1 M solution of methylthiol, HSCH3 since the initial 2 M solution has been doubled in volume during the titration. Since we know the pKb of its conjugate base SCH3– is 3.6, then we know the pKa of HSCH3 is 10.4. Therefore, at equilibrium: 10^–10.4 = (x)(x)/(1–x) Where x = [H+] = [SCH3–], at equilibrium. As the resultant value of x is negligibly small compared to 1, we ignore x in the denominator and say that x = [H+] = [SCH3–] = (10^–10.4)(0.5) giving 1 × 10–5.2. The answer –log (2 × 10^–5.2) would be correct if the resultant solution was 2 M in concentration. The answer –log (1 × 10^–1.8) is the result of using the pKb of the methylthiolate anion in the calculation rather than the pKa of HSCH3, and –log (1 × 10^–3.6) would be the result of a similar a calculation using this pKb value, while not taking the square root.

Answer 11

A. Addition of sodium acetate to a solution of acetic acid will cause the pH to increase due to the common ion effect. Sodium acetate is a basic compound, because acetate is the conjugate base of acetic acid, a weak acid ("the conjugate base of a weak acid acts as a base in water"). The addition of a base to any solution, whether it is buffered or not, will increase the pH. The answer is “increase due to the common ion effect.”

Answer 12

A. The resulting solution made from the combination of 50 mL of 1.0 M LiOH with 50 mL of 1.0 M HBr will be identical in all respects to 100 mL of 0.5 M LiBr. The best way to approach this question is to ignore OH– and H+, because they will neutralize each other when mixed. Thus, in this question we are mixing 50 mL of 1.0 M Li+ and 50 mL of 1.0 M Br–. Since the solution volume is doubling, the concentration of each ion is expected to decrease by a factor of 2. Therefore, the correct choice is “0.5 M LiBr”.

Answer 13

D. A titration of HF with NaOH will show an equivalence point above pH = 7. HF is the only weak acid of the hydrogen halide series, thus, when titrated with a strong base, it has a basic equivalence point. All the other choices have equivalence points at or very near pH = 7.

Answer 14

``` B. Enough HF (Ka = 7.4 × 10–4) is added to water to create a pH = 2.1 solution. The addition of PbF2 would have the least impact on the pH of this solution. ``` Both NaF and NH4F are soluble salts, and will decrease the solubility (acidity) of HF due to the common ion effect, so both of these answer choices can be eliminated. Na2CO3 is a soluble salt and CO32– is the conjugate base of a weak acid, so it will act to increase the pH. PbF2 is essentially insoluble in water, and therefore will have the least effect on the pH, making it the best answer.

Answer 15

B. The correct statement concerning 50 mL of basic solution composed of 0.2 M NaH2PO4 and 0.2 M Na2HPO4 is the pH of the solution will not change with the addition of 20 mL of water. This solution is a buffer. The pH of a buffer does not change with the addition or removal of water. see pics 6/10/21 for what I think the dissociation of each is

Answer 16

For the first question, when the base is titrated to its endpoint, it is all converted to the conjugate acid, HSCH3. For a pair of conjugates, pKa + pKb = 14. Therefore, the pKa of HSCH3 must be equal to 10.4. Since pKa = -log(Ka), you can put a negative sign in front of the pKa and turn it into an exponent to get Ka = 10^-10.4. Once you have the Ka of the weak acid, there are two equivalent ways to get the pH of a weak acid solution - use whichever is easier for you. One is to set up an ICE table, keeping in mind that Ka is an equilibrium constant for acid dissociation and is equal to [H+]*[A-]/[HA]. The other is to use the equation pH = -1/2*log(Ka*[WA]). For this question, be careful with the concentration - we doubled the volume when we added the HCl solution, so [HSCH3] = 1M. Then, pH = -1/2*log(10^-10.4*(1)), or (-1/2)(-10.4). The pH of the solution should be 5.2, which makes sense since it is a solution of a weak acid.

Answer 17

For the second question, keep track of how much H+ there is to neutralize. The 2M H2SO4 makes 4M H+ since it is diprotic, the 0.5M HI makes 0.5M H+, and the 3M HNO3 makes 3M H+, so there is 7.5M H+ total. Since there is 1 liter of acidic solution, that means there is 7.5 mol of H+ to neutralize, and we need 7.5 mol OH- ions. NaOH dissociates completely, so we can treat it as an equivalent of OH-. 7.5 mol OH- / (5 mol/L OH-) = 1.5 L, so we need 1.5 L of the 5M NaOH solution.

*High Yield GChem: Acids and Bases Flashcards

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