Reactions and Equilibrium - AP Chemistry

The question states that the patient has low levels of carbon dioxide. If we look at the given reaction, we will notice that the reaction will compensate for this by shifting the reaction equilibrium to the left. This phenomenon is called Le Chatelier’s principle and occurs to maintain the equilibrium of the reaction; therefore, the reaction will create more carbon dioxide by utilizing bicarbonate and hydrogen ions in the blood. A decrease in hydrogen ion concentration in blood will increase the pH and cause alkalosis (basicity in the blood). Since carbon dioxide is the cause of alkalosis, this patient will experience respiratory alkalosis. If he experienced alkalosis due to a change in bicarbonate ion concentration, the patient will have metabolic alkalosis.

The ratio of carbonic acid to bicarbonate will stay the same because both will be used in equal amounts (1:1 ratio) to produce carbon dioxide. Increasing respiratory rate, or hyperventilation, will result in an increase in the amount of carbon dioxide expelled by the patient; this will decrease the carbon dioxide concentration in the blood and will worsen the respiratory alkalosis. Recall that we are utilizing the bicarbonate ion (in conjunction with hydrogen ions) to create carbonic acid. The carbonic acid will be further broken down to replenish the carbon dioxide. A decrease in the bicarbonate concentration will slow down this process.

One way to make a buffer is by adding equal amounts of a weak acid to its weak conjugate base. For example, you can add 1M acetic acid to 1M acetate to create a buffer solution (note that both acetic acid and its conjugate base (acetate) are weak). However, when using this method you have to remember that the desired pH of the buffer solution has to equal the pKa of the weak acid. The question states that the pKa of the acid is 5.9 and the desired pH of the buffer is 3.5; therefore, it is not possible to make the buffer with the given acid. The researcher would have to find another acid that has a pKa near 3.5.

When performing acid/base calculations, the Henderson-Hasselbalch equation is useful:

To use this equation we need to convert the to the , and can use the following definition to do so:

Since an individual will begin experiencing acidosis when the blood falls below 7.35 we can use 7.35 as the pH in the Henderson-Hasselbalch equation. Adding in the of 6.10 calculated from the gives the equation below:

Plug in known values and solve:

Correct Answer:

The answer is unitless because all units cancel out.

For an oxyacid, the acid strength increases as the number of oxygens increase.

Acid strength increases going across a period.

pOH would be 4 (use –log \[OH–\]) and pH would be 14–pOH = 14 – 4 = 10

\[OH-\] = 4 X 10-6

pOH = 5.4 — use –log \[OH–\] to find pOH

pH = 14– pOH = 8.6

The pH of a solution is determined by taking the negative log of the concentration of hydrogen ions in solution. HCl is strong acid so it completely dissociates in solution. So adding .0001 M HCl is the same as saying that 1 *10-4 moles of H+ ions have been added to solution. The -log\[.0001\] =4, so the pH of the solution =4.

\[H+\] = 3.2 X 10-7 mol

pH = -log \[H+\]

= 7 - log 3.2

= 7 -.505

=6.495

pOH = 14 - pH

= 7.5

This question asks for the solution with the greatest number H+ ions, we can also approach this problem as if we are looking for the solution with the lowest pH. HCl is strong acid, therefore it will dissociate completely in solution. So the solution containing 0.010 mL of HCL contains 0.010 mL of H+ ions in solution, in addition to the H+ ions that are already in solution due to the auto-ionization of water. The only other solution that could have a pH less than 7 would be the one with 0.010 mL of CH3OH in excess water, becasue CH3OH is very slightly acidic. But since it is compared with an equal volume of HCl which is strong acid, it can be said that the most H+ ions will be found in the solution containing a small amount of strong acid, HCl.

Calcium carbonate is a base since it's the salt of a strong base (NaOH) and a weak acid (carbonic acid). The only basic pH on the list is 8, making it the correct answer.

The –log of the pKa will give you the Ka, so take the –log (6.4), which gives you approximately 4 * 10–7. The Ka expression is set up with products over reactants (hydrogen carbonate ion * hydrogen ion/carbonic acid). The undissociated carbonic acid is 0.001M, and you should use the variable 'x' to account for how much it dissociates and how many of the ions are produced. Ka = 4 * 10–7 = x2/0.001 ends up being your Ka expression, if you assume x is negligible compared to the original concentration of 0.001. Solving for x, you get 2 * 10–5. This is the hydrogen ion concentration. pH = –log (hydrogen ion concentration), so pH = –long(2 * 10–5), which is approximately 4.7.

The expression for Ksp is Ksp = \[Mg2+\]\[OH–\]2.

Thus, \[OH–\] = √(1.2 * 10–11)/(1.2 * 10–5)

\[OH–\] = 1 * 10–3

Thus, pH = –log(1 X 10–3) = 3

pH = 14 – 3 = 11

The Ka = \[H+\]\[F–\]/\[HF\]. The beginning concentration of HF is given as 0.05, and we can use x as the variable that accounts for how much of the HF is lost, along with how much of the H+ and F– are formed.

Thus, Ka = _x_2/.05 if you use the approximation that x is negligible compared to the starting concentration of HF.

Solving for x, x = 1 * 10–3. This is the H+ ion concentration.

pH = –log(H+) = –log(1.0 * 10–3)) = 3

NaOH is a base, so that won't produce an acidic solution. Of the remaining acids, HCl and HI are strong acids, and HF is weak. HI is at a higher molarity, so it will produce the most acidic solution.

Consider the reaction of and :

Now we will calculate the moles of in the solution prior to adding base.

We will then calculate the amount of moles of that react with the base.

We will then calculate the remaining moles of :

We will then calculate the new concentration of sulfuric acid:

Sulfuric acid is a diprotic acid, so the hydrogen ion concentration is 0.76 M.

First, we will calculate as follows:

Now we will calculate the pOH:

In solution , so .

The pH of the solution is 10.2, and we know:

And since , .

This question is simply testing your memorization of strong and weak acids. Of the list, you should recognize that nitric acid is the only strong acid, and the rest of the choices are weak.