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I'm really stuck on this question:

What volume of $0.0712 \mathrm{~mol~dm}^{–3}$ ammonium bromide solution will be required to precipitate all of the mercury(II) ions from $33.6 \mathrm{~cm}^3$ of $0.144\mathrm{~mol~dm}^{–3}$ mercury(II) nitrate solution? The ionic equation for the reaction is: $$\ce{Hg^{2+}(aq) + 2Br–(aq) -> HgBr2(s)}$$

The answer given is $136\mathrm{~mL}$.

What I have done so far:

I have figured out the number of moles of mercury(II) nitrate which is $0.05\ \mathrm{mol}$, and then used this to work out the number of mercury(II) ions ($3.01\times10^{22}$). I am unsure where to go from here, would I be required to multiply this value by two to get bromide ions?

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So, you have the total moles of Hg ions in solution.

Some question to get you going:

  1. According to your stoichiometry, how many moles of Br$^-$(aq) ions do you need to react with with one mole of Hg ions?
  2. Using that same ratio, how many moles of Br$^-$(aq) ions do you need to react with all of the Hg ions
  3. How many moles of Br$^-$(aq) does one mole of NH$_4$Br provide? How many total moles of NH$_4$Br do you need to provide all the moles of Br$^-$(aq) you need?
  4. Now that you know how many total moles of NH$_4$Br you need, how can you translate this amount to the total volume required given your concentration of $\frac{0.0712\text{ mol NH}_4\text{Br}}{1 \text{dm}^3}$
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