(2)

First NaOH reacts with HCl. Point B indicates complete neutralization of HCl. 2 mL of NaOH is used to neutralize HCl.

When NaOH completely neutralizes HCl:

$\text{mmol of NaOH}=\text{mmol of HCl}$

$M_{\text{NaOH}}\times V_{\text{NaOH}}\text{(that reacts with HCl)}=M_{\text{HCl}}\times V_{\text{HCl}}$

$0.1\times 2=M_{\text{HCl}}\times 40$

$M_{\text{HCl}}=\text{0.005\hspace{0.05em}M}$

When HCl is completely neutralized, then NaOH reacts with ${\mathrm{CH}}_3\mathrm{COOH}$.  Thus, (5−2) mL = 3 mL of NaOH reacts with ${\mathrm{CH}}_3\mathrm{COOH}$.  Point C indicates complete neutralization of ${\mathrm{CH}}_3\mathrm{COOH}$.

When NaOH completely neutralizes ${\mathrm{CH}}_3\mathrm{COOH}$:

$\text{mmol of NaOH}={\text{mmol of CH}}_3\text{COOH}$

$M_{NaOH}\times V_{NaOH}$ (that reacts with ${\mathrm{CH}}_3\mathrm{COOH}$)

$=M_{C{H}_{\mathit{3}}COOH}\times V_{C{H}_{\mathit{3}}COOH}$

$0.1\times 3$$=M_{C{H}_{3}COOH}\times 40$

$M_{C{H}_{3}COOH} =\text{0.0075\hspace{0.05em}M}$

(3)

Rate of corrosion is generally more in acidic medium.