<p>Please help me with acids and bases because I'm still not grasping the whole concept. can someone explain the reasoning behind the questions below:
1) Which of the following reagents could not be added together to make a buffer solution?
a) NaOH(aq) and CH3COOH(aq)
b) NaCH3COO and CH3COOH
c) NaOH and NaCH3 (answer)
d) NH4Cl and NH3</p>
<p>2) Which salt does not form an acidic solution in water?
a) MgCl2
b) Na2CO3 (answer)
c) FeCl3
d) NH4NO3</p>
<p>1) Buffers = weak acids or bases mixed with their conjugates.</p>
<p>2) Na+ is a spectator ion, so you can ignore it. The carbonate ion is basic so the salt obviously doesn't form an acidic solution in water.</p>
<p>Na2CO3 + H20 ---> NaOH (strong base) + H2CO3 (weak acid)
because you have a strong base and a weak acid, the solution is basic.</p>
<p>To expand on concrete's answer
1) A buffer solution contains a mixture of a weak acid/weak base conjugate pair, or a solution that will result in a conjugate pair. For choice (a) NaOH when added to acetic acid will convert it to the acetate anion. As long as the amount of NaOH added is low enough to create a mixture of acetic acid/acetate, you will have a buffer solution. (Too much converts it all to acetate - no buffering ability.) Choices (b) and (d) are already conj. mixtures. Choice (c) must have a typo, because NaCH3 isn't a real compound.
2) Acid/base properties of salts are evaluted by looking at the cation and anion separately. Cations can be thought of as conjugates of their respective bases. Anions can be though of as conjugates of their respective acids.
a) MgCl2_______<strong><em>Mg2+ conj of Mg(OH)2</em></strong><strong><em>Cl- conj of HCl
b) Na2CO3 (answer)</em></strong>Na+ conj of NaOH _______<strong><em>CO3 2- conj of H2CO3
c) FeCl3</em></strong>_<strong><em>Fe3+ conj of Fe(OH)3</em></strong><strong><em>Cl- conj of HCl
d) NH4NO3</em></strong><strong><em>NH4+ conj of NH4OH</em></strong>______NO3 1- conj of HNO3
The conjugate of a weak acid is a weak base; the conj of a weak base is a weak acid. The conjugate of a strong acid or base is neutral. Thus,
a) Mg2+ is acidic, Cl- is neutral
b) Na+ is neutral, CO3 2- is basic
c) Fe3+ is acidic, Cl- is neutral
d) NH4+ is acidic, NO3 1- is neutral</p>
<p>If something like NH4F were given (NH4+ is acidic, F- is basic) you would have to compare the Ka and Kb to decide.</p>
<p>What about these?</p>
<p>Neutralization of 500 ml of 2-molar NaOH requires the smallest volume of which of the folloiwng?
a) 1 M CH3COOH
b) 1 M H2SO4 (answer)
c) 1 M HCl
d) 1 M NH3
e) 0.1 M H2SO4</p>
<p>A solution is made by adding 5.6 grams of KOH (molar mass 56 grams) to enough water to make 1.0 liter of soltion. What is the approx. pH of the resulting solution?
a) 1
b) 3
c) 7
d) 9
e) 13 (answer)</p>
<p>The first question assumes full dissociation of strong acids. In other words, pick the highest concentration of the acid that produces the most protons (1M H2SO4 produces 2 protons per molecule, which is more than any of the others. Also, note that NH3 is a base, so you should eliminate it right off the bat).</p>
<p>Second question is really easy. 5.6g of KOH is 0.1 mol. If in 1 liter of solution, that's a 0.1M solution. KOH is a strong base, and under the assumption that it fully dissociates, that means the OH- concentration is 0.1M.</p>
<p>-log[OH-] gives you the pOH, which is 1. Since pH=14-pOH, the pH is 13.</p>
<p>Maybe this is picky, but the first question doesn't require you to assume full dissociation. Neutralizing 500 mL of 2-molar NaOH requires 1 mole of H+. The source (weak or strong acid) doesn't matter, because the OH- will pull any available acidic protons off any available acid. If in choice (b) 1 M H2SO4 was replaced with 1 M H2C6H6O6 (diprotic weak acid) the answer would still be (b).</p>
<p>Ah, you're right, my bad gfaith.</p>
<p>Whoa I just realized this is the AP prep forum.</p>
<p>I hope you (the OP) are not asking these questions in preparation for the AP test in May? This is basic stuff (or acidic... heh, get it?) that you should really know at this point.</p>
<p>@GoldShadow</p>
<p>For a problem like this, I would not waste time calculating anything. I would just choose "E" knowing that KOH completely disassociates and 5.6 grams of it is A LOT.</p>
<p>But your method is nonetheless correct.</p>
<p>Well the truth is, even if you were to calculate it on the actual test, it shouldn't take more than a few seconds to figure out that the pH is 13... regardless, you're right, it's easier and quicker to just look at it and tell.</p>
<p>No, I'm not taking the Ap test.</p>
<p>"Maybe this is picky, but the first question doesn't require you to assume full dissociation. Neutralizing 500 mL of 2-molar NaOH requires 1 mole of H+. The source (weak or strong acid) doesn't matter, because the OH- will pull any available acidic protons off any available acid. If in choice (b) 1 M H2SO4 was replaced with 1 M H2C6H6O6 (diprotic weak acid) the answer would still be (b)."</p>
<p>How did you come up with the fact that it takes 1 mole of H+ to neutralize the solution?</p>
<p>It's 500 mL of a 2M solution of KOH. That means 1 mol of OH- is present and requires 1 mol of H+ to neutralize it.</p>