In this question, we're presented with a variety of acid/base pairs and we're asked to identify which one could act as a buffer.

Remember that a buffer is a pair of acid and its conjugate base that acts to resist substantial changes in pH. In order for a buffer to work, the acid base pair needs to exist in equilibrium. This way, when the pH of the solution changes, the equilibrium of the acid/base reaction will shift, such that the pH will not change drastically.

To have an acid/base pair in equilibrium, we'll need to look for a pair that contains a weak acid. Acids like $\displaystyle HCl$ and $\displaystyle H_{2}SO_{4}$ are so strong that they will dissociate completely. Of the answer choices shown, only the carbonic acid/bicarbonate system ($\displaystyle H_{2}CO_{3}$ and $\displaystyle HCO_{3}^{-}$) exists in equilibrium. Thus, this is the correct answer.

At the half end point, the $\displaystyle pH=pKa$. This can be determined by the Henderson-Hasselbalch equation if it is not clear.

Since the endpoint of the titration is that there are 10 mL of 0.1 M NaOH added, that means that there are 0.001 moles of acetic acid. 

When 5 mL of NaOH is added, there are 0.0005 moles of acetic acid and 0.0005 moles of acetate formed. 

$\displaystyle pH = pKa + log \frac{base}{acid} = pKa + log \frac{0.0005}{0.0005 }$

$\displaystyle log(1) = 1$ 

Therefore pH= pKa 

For all the other options there is no ammonium leftover with which to serve as the weak acid in the buffer system, the ammonium is all used up and converted to ammonia. However in the correct answer choice, there is enough ammonium leftover after the reaction with the sodium hydroxide. 

Since acetic acid is a weak acid, it has a Ka that is rather small, we have to do a RICE table to determine the equilibrium amount of hydronium, H₃O⁺ to then determine the pH.  

R     $\displaystyle CH_{}3COOH (aq) + H_{}2O (l) \rightarrow CH_{}3COO^{}- (aq) + H_{}3O^{}+ (aq)$

I           0.1 M              -                 0                      0

C           -x                                    +x                   +x

E          0.1 -x                                 x                      x

So first we need to change our pKa to a Ka

where $\displaystyle Ka = 10^{}-^{^{^{}}}pKa$ therefore

$\displaystyle Ka= 10^{}-^{}4^{}.^{}7^{}5 = 1.78 x 10^{}-^{}5$  

^{$\displaystyle 1.78x10^{}-5$}  = $\displaystyle \frac{[CH_{}3COO^{}-][H_{}3O^{}+]}{[CH_{}3COOH]}$ = $\displaystyle \frac{x*x}{(0.1-x)}$

If we assume that x is very small compared to 0.1...

$\displaystyle 1.78x10^{}-5= \frac{x*x}{0.1}$

Where $\displaystyle x = 0.00133$

(note: when solving using the quadratic we come up with the same answer) 

So if $\displaystyle x= 0.00133 = [H_{}3O^{}+]$

$\displaystyle pH = -log [H_{}3O^{}+] =-log(0.0013) = 2.87$

These answers are correct because the two components needed to create a buffer solution are a weak acid and its conjugate base, or a weak base and its conjugate acid. In these cases, the first reaction to occur upon addition of the strong acid is the formation of the conjugate acid, acetic acid.

$\displaystyle HCl + CH_3COO^- \rightarrow CH_3COOH$

If the amount of initial $\displaystyle CH_3COO^-$ is greater than HCl, then we will have some $\displaystyle CH_3COO^-$ left over to act as a buffer with the created conjugate acid. This can be through a greater volume, or through a higher concentration as shown in the correct answers. 

A buffer solution is formed from the equilibrium of a weak acid and its conjugate base, or from a weak base and its conjugate acid. It's ability to "buffer" the pH or keep it from changing in large amounts in from the switching between these two forms weak and its conjugate. 

First to go through why the other ones are wrong:

Strong base + strong acid neutralizes and does not form a buffer solution

Strong base + weak base does not form a buffer - would need an acid

Strong base + weak acid  = all weak acid converted to conjugate base

The correct answer is: 

Strong base + weak acid = half converted to conjugate base with half leftover as weak acid, with all the components for a buffer 

Start by writing the chemical equation for this acid-base reaction.

$\displaystyle \text{HCl}(aq)+\text{NaOH}(aq)\rightarrow\text{H}_2\text{O}(l)+\text{NaCl}(aq)$

Next, find the number of moles of $\displaystyle \text{NaOH}$ that was added.

$\displaystyle 17.08\text{mL NaOH}\cdot\frac{1\text{L}}{1000\text{mL}}\cdot\frac{0.200\text{ moles NaOH}}{\text{L}}=0.003416\text{ moles of NaOH}$

At the equivalence point of a titration, the moles of acid and the moles of base are the same.

$\displaystyle \text{Moles of HCl}=\text{Moles of NaOH}=0.003416$

Finally, find the molarity of the solution.

$\displaystyle \text{Molarity}=\frac{\text{moles of solute}}{\text{liters of solution}}=\frac{0.003416}{15.0\times10^{-3}}=0.228M$

Make sure to round the answer to three significant figures.

Start by finding the initial number of moles of sodium hydroxide by using the given volume and molarity. Since sodium hydroxide is a strong base, it will dissociate completely into $\displaystyle \text{OH}^-$.

$\displaystyle \text{Initial moles of NaOH}=\text{Initial moles of OH}^-=0.075\text{L}(\frac{0.100\text{ mol}}{\text{L}})=0.0075\text{ moles}$

Next, find the number of moles of the hydrochloric acid added by using the given volume and molarity. Since hydrochloric acid is a strong acid, all of the $\displaystyle \text{H}_3\text{O}^+$ will come from the acid.

$\displaystyle \text{Moles of HCl added}=\text{Moles of H}_3\text{O}^+=0.045\text{L}(\frac{0.150\text{ moles}}{\text{L}})=0.00675\text{ moles}$

Now, as the acid is added, it will neutralize some of the base as shown by the following equation:

$\displaystyle \text{OH}^-(aq)+\text{H}_3\text{O}^+(aq)\rightarrow 2\text{H}_2\text{O}(l)$

Since the reactants are found in a $\displaystyle 1:1$ ratio, subtract the number of moles of acid added from the number of initial moles of base to get the amount of remaining base.

$\displaystyle \text{Remaining moles of NaOH}=0.0075-0.00675=7.5\times10^{-4}\text{ moles of NaOH}$

Now, divide this number of moles by the total volume of the solution to find the concentration of the base.

$\displaystyle [\text{OH}]^-=\frac{7.5\times 10^{-4}\text{ moles}}{0.075\text{L}+0.045\text{L}}=0.00625\text{ M}$

Next, find the $\displaystyle \text{pOH}$ value.

$\displaystyle \text{pOH}=-\log(0.00625)=2.204$

Finally, find the $\displaystyle \text{pH}$ value.

$\displaystyle \text{pH}=14-\text{pOH}$

$\displaystyle \text{pH}=14-2.204=11.80$

Your answer should have $\displaystyle 3$ significant figures.

$\displaystyle \text{pH}=11.8$

Determine the moles of $\displaystyle NaOH$ used:

$\displaystyle 0.035\ L\times \frac{0.025\ moles}{L}=8.75\times10^{-4}\ moles\ NaOH$

At the equivalence point of the titration:

$\displaystyle moles\ of\ NaOH=moles\ of\ HCl=8.75\times10^{-4}\ moles$

Convert the moles of $\displaystyle HCl$ to liters:

$\displaystyle \frac{L}{0.012\ moles}\times8.75\times10^{-4}\ moles=0.072\ L\ of HCl$

Convert the liters to milliliters:

$\displaystyle \frac{1\ mL}{10^{-3}L}\times0.072\ L=72.0\ mL\ HCl$