# How many sodium ions is each oxygen atom attracted to in aqueous sodium chloride? [closed]

When $$\ce{NaCl}$$ salt is dissolved in water the $$\ce{Cl-}$$ and $$\ce{Na+}$$ ions and the polar $$\ce{H2O}$$ molecules are attracted to each other such that each $$\ce{Na+}$$ ion attracts several oxygen atoms and each $$\ce{Cl-}$$ ion attracts several hydrogen atoms.

Solubility depends on temperature but around room temperature the solubility of $$\ce{NaCl}$$ in $$\ce{H2O}$$ is around $$100$$ g $$\ce{H2O}$$ to $$\pu{35 g}$$ $$\ce{NaCl}$$. This works out at a ($$\ce{H2O}:\ce{NaCl}$$) ratio of around ($$9:1$$) for a saturated solution.

So each $$\ce{Na+}$$ ion is attracted to up to around $$9$$ oxygen atoms depending on temperature. But is each oxygen atom only attracted to the one $$\ce{Na+}$$ ion at a time or are they simultaneously attracted to multiple $$\ce{Na+}$$ ions creating a kind of a web throughout the solution?

Every diagram I can find online suggests only one but I would like confirmation on this.

• You're asking about mechanism of NaCl dissolution, or what? Either way you're wrong. Commented May 30, 2020 at 17:00

You can make a quick estimate yourself based on observable bulk physical properties:

According to “Aqueous Solubility of Inorganic Compounds at Various Temperatures”, in CRC Handbook of Chemistry and Physics, 97th Edition (2016), William M. Haynes, ed., CRC Press/Taylor and Francis, Boca Raton, FL., the solubility of $$\ce{NaCl}$$ at $$20\ \mathrm{^\circ C}$$ is $$w_2=26.41\ \%$$ This solubility value is expressed as mass percent of solute, where $$w_2=\frac{m_2}{m_1+m_2}$$ and $$m_2$$ is the mass of solute and $$m_1$$ the mass of water. Therefore $$m_2=\frac{-w_2\cdot m_1}{w_2-1}$$ For example for a mass of water of $$m_1=100.0\ \mathrm g$$ \begin{align}m_2&=\frac{-w_2\cdot m_1}{w_2-1}\\[6pt] &=\frac{-0.2641\times100.0\ \mathrm g}{0.2641-1}\\[6pt] &=35.89\ \mathrm g\end{align} I.e., $$100.0\ \mathrm g$$ of water can dissolve $$35.89\ \mathrm g$$ of $$\ce{NaCl}$$.

Furthermore, you know that $$M(\ce{NaCl})=58.43977\ \mathrm{g\ mol^{-1}}$$ and $$M(\ce{H2O})=18.01500\ \mathrm{g\ mol^{-1}}$$. Thus, the molar ratio of water to $$\ce{NaCl}$$ is \begin{align}r&=\frac{n(\ce{H2O})}{n(\ce{NaCl})}\\[6pt] &=\frac{m_1/M(\ce{H2O})}{m_2/M(\ce{NaCl})}\\[6pt] &=\frac{100.0\ \mathrm g/18.01500\ \mathrm{g\ mol^{-1}}}{35.89\ \mathrm g/58.43977\ \mathrm{g\ mol^{-1}}}\\[6pt] &=\frac{5.551\ \mathrm{mol}}{0.6141\ \mathrm{mol}}\\[6pt] &=9.039\end{align}

This means that nine water molecules are needed to dissolve one $$\ce{Na+}$$ ion and one $$\ce{Cl-}$$ ion. Therefore, you may assume that the solvation process is not a simple one to one pairing.

When NaCl is dissolved in water, the sodium ions are surrounded by several water molecules which are all oriented with their oxygen atoms touching the sodium ion. But this number may fluctuate, as these water molecules are not chemically attached by covalent bonds and they are often bumped and even ejected by other water molecules due to the Brownian movement. The same thing happens to chloride ions which are surrounded by several water molecules even though they are not oriented like around the sodium ions. Anyway the ions are independent and can move separately in the solution. It is not a one to one pairing, as you think.

• Also search solvation number and hydration number.
– ACR
Commented May 30, 2020 at 17:50