Question

35.0 mL of 0.255 M nitric acid is added to 45.0 mL of 0.328 M Mg(NO3)2. What is the concentration of nitrate ion in the final solution?

Answer 1

there no reaction and a reaction isn't part of the question anyways find the moles of nitric acid we have \(0.035 L \cdot \frac{0.255 \mathrm{molHNO}_{3}}{1 L}=0.008925 \mathrm{molHN} O_{3}\)

within that solution, we have 0.008925 moles of HNO3. We need to find how many moles of NO3 is in that solution. according to the chemical formula, there is one mole of NO3 for every mole of HNO3. so from that solution we have \(0.008925 \mathrm{~mol}\) of nitrate ion(NO3-) find the moles of \(\mathrm{Mg}(\mathrm{NO} 3) 2\) we have

$$ 0.045 L \cdot \frac{0.328 \mathrm{molMg}\left(\mathrm{NO}_{3}\right)_{2}}{1 L}=0.01476 \mathrm{molMg}\left(\mathrm{NO}_{3}\right)_{2} $$

in this compound, for every mole of \(\mathrm{Mg}(\mathrm{NO} 3) 2\), there are twomoles of \(\mathrm{NO} 3\) so from this solution we have \(0.02952 \mathrm{~mol}\) of nitrate ion(NO3-)

add them up and find their new \(\mathrm{M}\) \(\mathrm{NO}_{3}^{-}\)

$$ \frac{\left(0.02952 \mathrm{~mol} N O_{3}^{-}+0.008925 \mathrm{~mol} N O_{3}^{-}\right)}{0.035 L+0.045 L}=\frac{0.4805625 \mathrm{~mol} N O_{3}^{-}}{\mid L}=0.4805625 M N O_{3}^{-}=0.481 M N O_{3}^{-} $$

the concetration of the new solution is \(0.481 \mathrm{M}^{N} \mathrm{O}_{3}^{-}\) thats with sig figs and all