In this question, we're given the concentrations of both a weak acid and its conjugate base in solution. We're also provided with the acid-dissociation constant for this acid, and we're asked to find the pH of the solution.

The easiest way to go about solving this problem is to first convert the acid-dissociation constant given into pKa. Then, we can use the Henderson-Hasselbalch equation to solve for the pH of this solution.

\(\displaystyle pK_{a}=-log(K_{a})\)

\(\displaystyle pK_{a}=-log(6.76\cdot 10^{-4})\)

\(\displaystyle pK_{a}=3.17\)

Now that we have the pKa, we can use it, along with the concentrations of the weak acid and its conjugate base, in order to solve for the pH of the solution.

\(\displaystyle pH=pKa+log(\frac{base}{acid})\)

\(\displaystyle pH=3.17+log(\frac{2.5}{3.0})\)

\(\displaystyle pH=3.09\)

The Henderson-Hasselbach equation can be used to find the answer.

\(\displaystyle pH = pKa + log \left(\frac{[A^{-}]}{[HA]}\right)\)

\(\displaystyle [A^{-}]\) represents the concentration of conjugate base, while \(\displaystyle [HA]\) represents the concentration of weak acid.

When we plug the given numbers into this equation, we get:

\(\displaystyle 7.4 = 4.91 + log \left(\frac{[A^{-}]}{[HA]}\right)\)

\(\displaystyle 2.49 = log \left(\frac{[A^{-}]}{[HA]}\right)\)

\(\displaystyle 10^{2.49} = \left(\frac{[A^{-}]}{[HA]}\right)\)

\(\displaystyle 309.03 = \left(\frac{[A^{-}]}{[HA]}\right)\)

To find the answer to this question, we must use the Henderson-Hasselbach equation:

\(\displaystyle pH = pKa + log\left(\frac{[A^{-}]}{[HA]}\right)\)

\(\displaystyle [A^{-}]\) refers to the concentration of base (in this case, ammonium chloride). \(\displaystyle [HA]\) refers to the concentration of weak acid (in this case, ammonia). We must then plug the given numbers into the equation and solve for pH.

\(\displaystyle pH = 9.248 + log\left(\frac{0.9}{0.7}\right)\)

\(\displaystyle pH = 9.248 + log(1.29)\)

\(\displaystyle pH = 9.248 + 0.11\)

\(\displaystyle pH = 9.358\)

In this question, we're given two different solutions. For each, we're told the volume as well as the concentration of acid within that solution. Then, we're told that the two solutions are mixed. Upon combining them, we're asked to determine what the pH of the resultant mixture will be.

To answer this question, the first thing we have to realize is that acetic acid and its conjugate base, acetate, can act as a buffer system. This means that it acts to resist dramatic changes in pH. Because we are told that acetic acid and its conjugate base are in equal amounts, we know that the pH of the starting solution is equal to the pKa of acetic acid.

\(\displaystyle CH_{3}COOH\rightleftharpoons CH_{3}COO^{-}+H^{+}\)

When \(\displaystyle HCl\) is added to the solution, we know that it will completely dissociate because it is a strong acid. This will, in turn, make the solution more acidic. The buffer system, however, will help to ensure that the pH does not drop too significantly, although we can still expect the pH to fall slightly below the pKa.

It's important to remember that when the two solutions are combined, the volume changes as well. Since each solution begins at \(\displaystyle 500\:mL\), combining them together will result in a final solution with twice the volume, or \(\displaystyle 1\:L\). The significance of this is that the molarity of acetic acid and acetate will be cut in half, each to a value of \(\displaystyle 1.25\:M\). And upon reacting with the extra protons produced by the \(\displaystyle HCl\), this will also change the concentration slightly.

In this new mixture, \(\displaystyle 0.1\:moles\) of the protons produced by \(\displaystyle HCl\) will combine with an equivalent \(\displaystyle 0.1\:moles\) of acetate to produce an additional \(\displaystyle 0.1\:moles\) of acetic acid. Hence, we can calculate the final molarity of the buffer components. Acetic acid will have a concentration of \(\displaystyle 1.35\:M\) and acetate will have a concentration of \(\displaystyle 1.15\:M\).

We now have what we need to calculate the pH of the mixture. Plugging these values into the Henderson-Hasselbalch equation will give us the answer.

\(\displaystyle pH=pKa+log(\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]})\)

\(\displaystyle pH=4.76+log(\frac{1.15\:M}{1.35\:M})=4.69\)

To answer this question, we can use the Henderson-Hasselbalch equation since we have a buffer system, a weak acid and its conjugate base. 

Acetic acid is our weak acid  and acetate is the conjugate base.

\(\displaystyle pH= pKa + log\frac{[base]}{[acid]} = pKa + log \frac{acetate}{aceticacid}\)

Since we are given volumes and concentration we need to figure out the molar ratio between acetate and acetic acid. We cannot just plug in the given concentrations.

\(\displaystyle molesacetic acid = 0.03 \frac{moles}{L} * \frac{1 L}{1000 mL}* 10 mL = 3x10^{-4}moles\)

\(\displaystyle moles of acetate = 0.025 \frac{moles}{L} * \frac{1L}{1000 mL} * 15 mL = 3.75x10^{-4} moles\)

\(\displaystyle pH = 4.75 + log \frac{3.75x10^{-4} moles}{3x10^{-4} moles}= 4.85\)

This pH makes sense because the number should be more basic. We have more moles of the base than we do of the acid. If we had more moles of acid we would expect the pH to be lower.

The answer is the acetate ion because at this point, there has not been enough hydrochloric acid added to make a lot of acetic acid. And the hydrochloric acid is immediately used up upon addition into the solution. The water is at a constant concentration so is not considered the predominant species. 

The first reaction that has to happen is the creation of \(\displaystyle NH_4^+\), ammonium.

\(\displaystyle H^+ + NH_3 \rightarrow NH_4^+\)

We need to determine how many moles of hydrogen ions that we have from the HCl, that is equal to the amount of \(\displaystyle NH_4^+\) that is formed.

\(\displaystyle moles = 0.04 \frac{moles}{L}* \frac{1L}{1000mL }*8 mL = 3.2x10^{^{-4}}moles\)

We need to determine the total amount of moles of ammonia, and then how many are remaining.

\(\displaystyle totalmoles= 0.8 \frac{moles}{L} * \frac{1L}{1000mL }*5 mL = 0.004 moles\)

\(\displaystyle remaining moles = 0.004 moles-3.2x10^{-4}moles=0.00368moles\)

Now we can use the Henderson-Hasselbalch equation to solve for the pH

\(\displaystyle pKa = pH + log \frac{base}{acid} = 9.26 + log \frac{3.2x10-4}{0.00368}= 8.2\)

The relevant chemical reaction here is: 

\(\displaystyle NH_4^+ + H_2O \Leftrightarrow NH_3 + H_3O^+\)

First we need to determine how many moles we have of \(\displaystyle NH_4^+\) and \(\displaystyle NH_3\)_. 

\(\displaystyle moles ammonium = 0.25 \frac{mole}{L} * \frac{1L}{1000 mL}* 6 mL = 0.0015 moles\)

\(\displaystyle moles ammonia = 0.34 \frac{moles}{L} * \frac{1 L}{1000mL}* 5 mL = 0.0017 moles\)

Now we can use the Henderson-Hasselbalch equation to solve for the pH.

\(\displaystyle pH = pKa + log \frac{base}{acid} = 9.26 + log \frac{0.0017}{0.0015}= 9.31\)

The first reaction that will happen is the formation of acetate from the reaction of the strong base with acetic acid:

\(\displaystyle OH_- + CH_3COOH \rightarrow CH_3COO^- + H_2O\)

How many moles of OH^- are reacting? Using the volume:

\(\displaystyle moles = 0.5 \frac{moles}{L} * \frac{1L}{1000 mL} * 10 mL = 0.005 moles\)

How many moles of CH₃COOH are reacting? Doing the same as above: 

\(\displaystyle moles = 0.5 \frac{moles}{L} * \frac{1L}{1000mL}* 25 mL = 0.0125 moles\)

So the amount of acetate, \(\displaystyle CH_3COO^-\)formed is equal to the number of moles of \(\displaystyle OH^-\) added. 

And to figure out how much acetic acid is leftover: 

moles of acetic acid - moles of \(\displaystyle OH^- = 0.0125-0.005 = 0.0075\) moles of acetic acid leftover

So now that we know how many moles we have of the weak acid and its conjugate base, we can use the Henderson-Hasselbalch equation: 

\(\displaystyle pH = pKa + log\frac{base}{acid} = 4.75 + log \frac{0.005}{0.0075} = 4.57\)

We need to use the Henderson-Hasselbalch equation to determine how many moles of ammonia we need.

First we need to determine the number of moles of ammonium we have: 

\(\displaystyle moles = 0.025 \frac{moles}{L}* \frac{1L}{1000mL}*45mL=0.001125 moles\)

When we plug that into the Henderson-Hasselbalch equation:

\(\displaystyle pH=pKa + log\frac{base}{acid}\)

In this case, we have the acid, and we are looking for the base

\(\displaystyle 7.34 = 9.26 + log\frac{x}{0.001125}\)

\(\displaystyle -1.92 = log\frac{x}{0.001125}\)

Raise both sides to the 10

\(\displaystyle 10^{-1.92}= \frac{x}{0.001125}\)

Cross multiply

\(\displaystyle x=1.35x10^{-5}\)moles of ammonia

Now we need to determine the volume of 0.1 M ammonia that gives us that number of moles:

\(\displaystyle mL = \frac{1000 mL}{L} *\frac{L}{0.1 moles} * 1.35x10^{-5}moles = 0.135 mL\)