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:

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

Determine the moles of $\displaystyle NaOH$ used:

$\displaystyle 0.042\ L\times \frac{0.0021\ moles}{L}=8.82\times10^{-5}\ moles\ NaOH$

At the equivalence point of the titration:

$\displaystyle moles\ of\ NaOH=moles\ of\ HCl=8.82\times10^{-5}\ moles$

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

$\displaystyle \frac{L}{0.014\ moles}\times8.82\times10^{-5}\ moles=0.0063\ L\ of HCl$

Convert the liters to milliliters:

$\displaystyle \frac{1\ mL}{10^{-3}L}\times0.0063\ L=6.3\ mL\ HCl$

Using the concentration of $\displaystyle NaOH$ as a conversion factor, we can convert the volume of $\displaystyle NaOH$ to moles by dimensional analysis:

$\displaystyle \frac{0.1\ moles}{L}\times0.032\ L=0.0032\ moles\ NaOH$

At the equivalence point of the titration:

$\displaystyle moles\ of\ NaOH=moles\ of\ CH_{3}COOH=0.0032\ moles$

$\displaystyle Molarity=\frac{0.0032\ moles}{0.025\ L}=0.13\ M$

$\displaystyle \frac{0.1\ moles}{L}\times0.015L=0.0015\ moles$

At the equivalence point of the titration:

$\displaystyle moles\ of\ NaOH=moles\ of\ CH_{3}COOH=0.0015\ moles$

Convert the moles of $\displaystyle CH_3COOH$ to grams:

$\displaystyle 0.0015\ moles\ CH_{3}COOH\times \frac{60.05\ g}{moles}=0.090\ g\ CH_{3}COOH$

Using the density of $\displaystyle CH_3COOH$ given, convert the grams to mL:

$\displaystyle 0.09\ g \times \frac{mL}{1.05g}=0.086\ mL\ CH_{3}COOH$

Determine the percentage of $\displaystyle CH_3COOH$ by dividing the volume of $\displaystyle CH_3COOH$ by the total volume of vinegar:

$\displaystyle \frac{0.086\ mL}{40.0\ mL}=0.21\%\ CH_{3}COOH\ in\ vinegar$

Using the concentration of $\displaystyle NaOH$ as a conversion factor, we can convert the volume of $\displaystyle NaOH$ to moles by dimensional analysis:

$\displaystyle \frac{0.1\ moles}{L}\times0.0267\ L=0.00267\ moles\ NaOH$

At the equivalence point of the titration:

$\displaystyle moles\ of\ NaOH=moles\ of\ CH_{3}COOH=0.00267\ moles$

$\displaystyle Molarity=\frac{0.0032\ moles}{0.0208\ L}=0.154\ M$

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

$\displaystyle \frac{2.00\ moles}{L}\times0.030L=0.06\ moles\ NaOH$

At the equivalence point of the titration:

$\displaystyle moles\ of\ NaOH=moles\ of\ CH_{3}COOH=0.06\ moles$

Convert the moles of $\displaystyle CH_3COOH$ to grams:

$\displaystyle 0.06\ moles\ CH_{3}COOH\times \frac{60.05\ g}{moles}=3.6\ g\ CH_{3}COOH$

Using the density of $\displaystyle CH_3COOH$ given, convert the grams to mL:

$\displaystyle 3.6\ g \times \frac{mL}{1.05g}=3.4\ mL\ CH_{3}COOH$

Determine the percentage of $\displaystyle CH_3COOH$ by dividing the volume of $\displaystyle CH_3COOH$ by the total volume of vinegar:

$\displaystyle \frac{3.4\ mL}{100.0\ mL}=3.4\%\ CH_{3}COOH\ in\ vinegar$