In this paper, the production of $\ce{SnO2}$ nanoparticles through the chemical co-precipiation method is described using $\ce{SnCl2·2H2O}$ as a precursor. According to their experimental section,

$\ce{SnO2}$ NPs were prepared by chemical co-precipitation method. 0.27 M $\ce{SnCl2⋅2H2O}$ was added to double destilled (DD) water and ethanol (2.2:1 v/v) in a 1000 mL beaker under continuous magnetic stirring for 60 min in order to obtain a homogeneous solution. 50 mL of $\ce{HCl}$ was added drop wise in order to reduce the precursor and the solution was kept on stirring for another 30 min to allow proper mixing resulting in a clear transparent solution. Later aqueous $\ce{NH3}$ was added dropwise till the pH of the solution became 8.5 and a milky white solution was obtained. The solution was kept on continuous stirring for another 3 h. After this the stirring was stopped and the beaker was kept undisturbed overnight in order to facilitate the particles to settle down completely.

The solution was then washed with DD water and ethanol multiple times so as to remove the residual chlorine content using 125 mm Whatman filter paper (Cat no. 1001 125). The resulting product was covered and left for drying overnight. The dried powder was then divided into three parts and annealed at 400 °C, 500 °C, and 600 °C named as Sn4, Sn5, and Sn6 respectively.

However, the paper focuses mainly in the characterization of the syntetized nanoparticles, and it doesn't describe in further detail the chemical reactions of the process. What would the role of the added $\ce{HCl}$ be? Why should the $\ce{NH3}$ be added, and why should it be added until an 8.5 pH is obtained? What would the chemical reaction of the process be? I would like to know the reason behind these steps in order to fully understand the process, unfortunately the paper doesn't cover them.

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    $\begingroup$ Some concepts are wrong in the paper. HCl is not there to reduce anything, it is there to prevent hydrolysis of tin chloride. Addition of ammonia is for controlled hydrolysis of the tin salt. The pH 8.5 must be empirical. They must have done a lot of hit and trials. Did you search other papers? $\endgroup$
    – ACR
    Commented Jul 28, 2023 at 12:39

1 Answer 1


Tin(II) chloride $\ce{SnCl2}$ is not perfectly soluble in pure water. It is only soluble in aqueous $\ce{HCl}$ solutions. In pure water $\ce{SnCl2}$ is partially hydrolyzed according to the equilibrium $$\ce{SnCl2 + 2 H2O <=> Sn(OH)2 + 2 HCl}$$ If enough $\ce{HCl}$ is present in solution, $\ce{SnCl2}$ gets dissolved without hydrolysis, and produces $\ce{Sn^{2+}}$ ions. But in the presence of ammonia, $\ce{HCl}$ is destroyed and the $\ce{Sn^{2+}}$ ion is precipitated as $\ce{Sn(OH)2}$ according to the irreversible reaction : $$\ce{Sn^{2+} + 2 NH3 + 2 H2O -> Sn(OH)2(s) + 2 NH4^+}$$ The trouble is that $\ce{Sn(OH)2}$ is easily oxidized by atmospheric oxygen. So in contact with air, $\ce{Sn(OH)2}$ is quickly oxidized into $\ce{SnO2}$ according to : $$\ce{2 Sn(OH)2 + O2 -> 2 SnO2 + 2 H2O}$$ This is why tin(II) chloride is easily transformed and oxidized into insoluble $\ce{SnO2}$.

  • $\begingroup$ Very helpful! One last thing, the paper specifies to use tin(II) chloride dihydrate, would it be irrelevant to use non-hydrated tin(II) chloride, as you did in the answer? $\endgroup$ Commented Jul 31, 2023 at 6:12
  • $\begingroup$ @User. No ! All I said about $\ce{SnCl2}$ can be also said about its hydrate $\ce{SnCl2·2H2O}$ or any other hydrate. I used the formula $\ce{SnCl2}$ in my answer because it is simpler than $\ce{SnCl2·2H2O}$ $\endgroup$
    – Maurice
    Commented Jul 31, 2023 at 16:26

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