Weigh a phosphate sample (possible components $\mathrm{Na_3PO_4 ~, Na_2HPO_4 ~, NaH_2PO_4 ~) ~ 0.7250 ~ g}$, dissolve it in water, add phenolphthalein indicator, and titrate with standard HCl solution (0.2000 mol/L) until the endpoint is reached, consuming 20.02 ml. For a sample of the same mass, using methyl orange as the indicator, titrate with the same concentration of HCl to the endpoint, consuming 42.04 ml. What is the percentage content of each component?($M_{\mathrm{Na_3PO_4}} = 163.9 \;, \; M_{\mathrm{Na_2HPO_4}} = 142.0 \;, \; M_{\mathrm{NaH_2PO_4}} = 120.0$)

Options:
A. $$\begin{aligned}&w_{\mathrm{Na_3PO_4}}(\:\%\:)\:=\:\frac{0.2300\times22.02\times163.9}{0.7250\times1000}\times100\%\:=\:82.52\%\\&w_{\mathrm{Na_2HPO_4}}(\:\%\:)\:=\:\frac{0.2000\times(\:42.\:04\:-5\times20.\:02\:)\times142.\:0}{0.\:7250\times1000}\times100\%\:=\:8.\:834\%\end{aligned}$$
B. $$\begin{aligned}&w_{\mathrm{Na_3PO_4}}(\:\%\:)\:=\:\frac{0.2600\times24.02\times163.9}{0.7250\times1000}\times100\%\:=\:89.52\%\\&w_{\mathrm{Na_2HPO_4}}(\:\%\:)\:=\:\frac{0.2000\times(\:42.\:04\:-9\times20.\:02\:)\times142.\:0}{0.\:7250\times1000}\times100\%\:=\:13.\:834\%\end{aligned}$$
C. $$\begin{aligned}&w_{\mathrm{Na_3PO_4}}(\:\%\:)\:=\:\frac{0.2400\times23.02\times163.9}{0.7250\times1000}\times100\%\:=\:80.52\%\\&w_{\mathrm{Na_2HPO_4}}(\:\%\:)\:=\:\frac{0.2000\times(\:42.\:04\:-8\times20.\:02\:)\times142.\:0}{0.\:7250\times1000}\times100\%\:=\:7.\:834\%\end{aligned}$$
D. $$\begin{aligned}&w_{\mathrm{Na_3PO_4}}(\:\%\:)\:=\:\frac{0.1900\times15.02\times163.9}{0.7250\times1000}\times100\%\:=\:95.52\%\\&w_{\mathrm{Na_2HPO_4}}(\:\%\:)\:=\:\frac{0.2000\times(\:42.\:04\:-4\times20.\:02\:)\times142.\:0}{0.\:7250\times1000}\times100\%\:=\:5.\:834\%\end{aligned}$$
E. $$\begin{aligned}&w_{\mathrm{Na_3PO_4}}(\:\%\:)\:=\:\frac{0.2800\times18.02\times163.9}{0.7250\times1000}\times100\%\:=\:85.67\%\\&w_{\mathrm{Na_2HPO_4}}(\:\%\:)\:=\:\frac{0.2000\times(\:42.\:04\:-1\times20.\:02\:)\times142.\:0}{0.\:7250\times1000}\times100\%\:=\:9.\:834\%\end{aligned}$$
F. $$\begin{aligned}&w_{\mathrm{Na_3PO_4}}(\:\%\:)\:=\:\frac{0.2100\times25.02\times163.9}{0.7250\times1000}\times100\%\:=\:75.52\%\\&w_{\mathrm{Na_2HPO_4}}(\:\%\:)\:=\:\frac{0.2000\times(\:42.\:04\:-3\times20.\:02\:)\times142.\:0}{0.\:7250\times1000}\times100\%\:=\:12.\:834\%\end{aligned}$$
G. $$\begin{aligned}&w_{\mathrm{Na_3PO_4}}(\:\%\:)\:=\:\frac{0.2500\times19.02\times163.9}{0.7250\times1000}\times100\%\:=\:88.52\%\\&w_{\mathrm{Na_2HPO_4}}(\:\%\:)\:=\:\frac{0.2000\times(\:42.\:04\:-6\times20.\:02\:)\times142.\:0}{0.\:7250\times1000}\times100\%\:=\:6.\:834\%\end{aligned}$$
H. $$\begin{aligned}&w_{\mathrm{Na_3PO_4}}(\:\%\:)\:=\:\frac{0.2000\times20.02\times163.9}{0.7250\times1000}\times100\%\:=\:90.52\%\\&w_{\mathrm{Na_2HPO_4}}(\:\%\:)\:=\:\frac{0.2000\times(\:42.\:04\:-2\times20.\:02\:)\times142.\:0}{0.\:7250\times1000}\times100\%\:=\:7.\:834\%\end{aligned}$$
I. $$\begin{aligned}&w_{\mathrm{Na_3PO_4}}(\:\%\:)\:=\:\frac{0.2200\times21.02\times163.9}{0.7250\times1000}\times100\%\:=\:87.52\%\\&w_{\mathrm{Na_2HPO_4}}(\:\%\:)\:=\:\frac{0.2000\times(\:42.\:04\:-7\times20.\:02\:)\times142.\:0}{0.\:7250\times1000}\times100\%\:=\:10.\:834\%\end{aligned}$$
J. $$\begin{aligned}&w_{\mathrm{Na_3PO_4}}(\:\%\:)\:=\:\frac{0.2700\times26.02\times163.9}{0.7250\times1000}\times100\%\:=\:83.52\%\\&w_{\mathrm{Na_2HPO_4}}(\:\%\:)\:=\:\frac{0.2000\times(\:42.\:04\:-10\times20.\:02\:)\times142.\:0}{0.\:7250\times1000}\times100\%\:=\:11.\:834\%\end{aligned}$$
Answer: H