Question

a 250.0ml solution of 0.200mol L-1 aluminum nitrate was mixed together with 200.0ml of 0.200mol L-1 barium nitrate. What is the concentration of nitrate ions in the resultant solution?

Answer 1

Moles of $NO^-_3$ in aluminum nitrate$=3$ times the moles of aluminum nitrate.(Chemical formula$\equiv Al(NO_3)_3$ )

Moles of $NO_3^-$ in aluminium nitrate solution$=3\times \frac{250}{1000}\times 0.200=0.15$

Moles of $NO_3^-$ in barium nitrate$=2$ times the moles of barium nitrate (Chemical formula$=Ba(NO_3)_2$ )

Moles of $NO_3^-$ in barium nitrate solution$=2\times \frac{200}{1000}\times 0.200= 0.08$

Total moles$=0.15+0.08=0.23$

Total volume$=200+250=450 \ ml=0.45 \ l$

Concentration in resultant solution$=\frac{0.23}{0.45}=0.51 \ mol/l$