# How many Cl and Na atoms per Water molecule when salt dissolves? [closed]

When salt dissolves in water the Cl and Na ions or drawn to the polar water molecules.

Is it a one to one pairing? As in one Cl to the hydrogen side and one Na to the oxygen side?

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

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. – M. Farooq May 30 at 17:50

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.

• Writing such a substantial answer to the question scoring −3 with four close-votes is a risky business. You never know until you try:) – andselisk Jun 1 at 9:51
• @andselisk ....and the question is now closed ;) – Nilay Ghosh Jun 1 at 9:59