Zn can reduce \(\mathrm{NO}_{3}^{-}\) to \(\mathrm{NH}_{3}(\mathrm{g})\) in basic solution. (The following equation is not balanced.) $$\begin{aligned}\mathrm{NO}_{3}^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s})+& \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \\\&\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{g})\end{aligned}$$ The \(\mathrm{NH}_{3}\) can be neutralized with an excess of \(\mathrm{HCl}(\mathrm{aq})\) Then, the unreacted HCl can be titrated with NaOH. In this way a quantitative determination of \(\mathrm{NO}_{3}^{-}\) can be achieved. A 25.00 mL sample of nitrate solution was treated with zinc in basic solution. The \(\mathrm{NH}_{3}(\mathrm{g})\) was passed into \(50.00 \mathrm{mL}\) of \(0.1500 \mathrm{M} \mathrm{HCl} .\) The excess \(\mathrm{HCl}\) required \(32.10 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for its titration. What was the \(\left[\mathrm{NO}_{3}\right]\) in the original sample?

Step by step solution

01

Determine the moles of NaOH used in the titration of excess HCl

The volume of \(0.1000 \,M\) \(NaOH\) used was \(32.10\, mL\), or \(32.10 \times10^{-3}\, L\). The molarity (M) is defined as moles/L, which we can rearrange to find the moles of \(NaOH\) as \(M \times \text{Volume in L}\), \(0.1000 \times 32.10 \times10^{-3} = 3.210 \times 10^{-3}\, \text{moles of NaOH}\).

02

Determine the moles of HCl that reacted with NH3

The moles of \(NaOH\) and \(HCl\) will be the same because \(NaOH\) and \(HCl\) react in a 1:1 ratio. Therefore, there were \(3.210 \times 10^{-3}\, moles\) of \(HCl\) unused. The initial amount of \(HCl\) was \(50.00 \times 10^{-3} \, L \times 0.1500 \, M = 7.500 \times 10^{-3}\, \text{moles}\). Thus, the moles of \(HCl\) that reacted with \(NH_{3}\) is \(7.500 \times 10^{-3} - 3.210 \times 10^{-3} = 4.290 \times 10^{-3}\, \text{moles}\).

03

Determine the concentration of \(NO_{3}^{-}\) in the original sample

Since \(NO_{3}^{-}\) was converted into \(NH_{3}\) in a 1:1 ratio, the moles of \(NH_{3}\) is also \(4.290 \times 10^{-3}\, \text{moles}\). Lastly, divide the number of moles of nitrate by the volume of the original nitrate solution to get the original concentration, \(4.290 \times 10^{-3} \, moles / (25.00 \times 10^{-3}) \, L = 0.172 \, M\).

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