A mixture of NaOH and  $N{a}_{2}C{O}_3$ required 25 mL of 0.1 M HCl using phenolphthalein as the indicator. However, the same amount of the mixture required 30 mL of 0.1 M HCl when methyl orange was used as the indicator. The simplest whole number molar ratio of NaOH and  $N{a}_{2}C{O}_3$ in the mixture is x : y, then find the value of x + y?

With phenolphthalein  $NaOH+HCl\to NaCl+H_{2}O$
  $$\begin{gathered} N{a}_{2}C{O}_3+HCl\to NaHC{O}_3+NaCl \\ \therefore {\text{m-moles\hspace{0.17em}of\hspace{0.17em}NaOH+m-moles\hspace{0.17em}of\hspace{0.17em}Na}}_{\text{2}}{\text{CO}}_{\text{3}} \end{gathered}$$
   = m-moles of HCl = 2.5
 Methyl orange gives the end point corresponding to the reactions:
  $$\begin{gathered} NaOH+HCl\to NaCl+H_{2}O \\ N{a}_{2}C{O}_3+2HCl\to 2NaCl+C{O}_2+H_{2}O \end{gathered}$$
  $\therefore m-molesofNaOH$   + m moles of  ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{×2}$
 = m- moles of HCl  $=30\times 0.1=3$
 m-moles of  $N{a}_{2}C{O}_3=0.5$
 m-moles of NaOH = 2.5 – 0.5 =2
 Ratio of m-moles of NaOH and  ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$
 $\text{=}\frac{2}{0.5}=4:1$