A 100 mL portion of 0.250 M  calcium nitrate solution is mixed with $400 \mathrm{mL}$ of  $0.100 \mathrm{M}$ nitric acid solution. What is the final concentration of the nitrate ion?

$\text{ Moles Ca(NO}{{\text{3)}}_{\text{2}}}_{ }\text{2}=\frac{100\times 0.250}{1000}$

$=0.025\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2>{\mathrm{Ca}}^{2++}$

$2{\mathrm{NO}}_3-{\text{ Moles NO}}_{\text{3}}\text{- = }2\times 0.025=0.05$

$\text{ Moles }{\mathrm{HNO}}_3=400\times 0.100/1000=0.04$

$\text{ Total moles }=0.05+0.04=0.09$

$\text{ Total volume }=500\mathrm{ml}=0.500\mathrm{LM}=0.09 /0.500 =0.18$

$\frac{100\times 0.25\times 2+400\times 0.1}{500}=0.18M$