# Density of a sodium hydroxide solution

The specific gravity of a $$50\%$$ aqueous solution $$\ce{NaOH}$$ is $$\pu{1.5298 g cm-3}.$$

To my understanding, the $$50\%$$ means $$50$$ mass fraction $$w,$$ i.e. $$\pu{500 g}$$ $$\ce{NaOH}$$ in $$\pu{500 g}$$ $$\ce{H2O}:$$

$$w(\ce{NaOH}) = \frac{m(\ce{NaOH})}{m(\ce{NaOH}) + m(\ce{H2O})} = \frac{\pu{500 g}}{\pu{500 g} + \pu{500 g}} = 0.5.\tag{1}$$

However, when I try to calculate the density of that solution I end up with

\begin{align} \rho(\ce{NaOH}) &= \pu{2.13 g cm-3}; \\ \rho(\ce{H2O}) &= \pu{1.00 g cm-3}. \end{align}

$$V = \frac{m(\ce{NaOH})}{\rho(\ce{NaOH})} + \frac{m(\ce{H2O})}{\rho(\ce{H2O})} = \pu{234.74 cm^3} + \pu{500 cm^3} = \pu{734.74 cm^3},\tag{2}$$

$$\rho_\text{sln} = \frac{\pu{1000 g}}{\pu{734.74 cm^3}} = \pu{1.361 g cm-3}.\tag{3}$$

What am I missing here?

• You don't calculate density. Aug 31 '20 at 10:21
• Densities are rarely, if ever, directly proportional to mass percentage of a solute. You need to determine the values experimentally.
– MaxW
Aug 31 '20 at 10:22
• Plotting the densities to mass percentage creates an almost linear function (in this case). It is considered proportional in many cases. $ϱ_{sol}=0.0106W_{NaOH} + 1.0034$ Aug 31 '20 at 10:26
• Thanks for the input so far. I'm aware that density needs to be measured. I'm thinking of mixing ethanol and water, where the volume of the solution will change. I was suspecting that I get the weight percentage part wrong. My calculations would yield that 1 L of 50% solution contains 680 g NaOH. According to the table I originally linked it would be 764 g / L. So basically my calculation of the solution is correct but I would need to look up the density in a table or use a correlation like @AndrewKovács suggested? Aug 31 '20 at 10:52
• You have added volumes. This is wrong. Volumes in mixtures are never additive. Masses are additive, but not volumes. $234$ mL pure NaOH +$500$ mL water does not give $734$ mL solution. It gives less. Aug 31 '20 at 11:38

What's your first miss is the definition of $$50\%(w/w)$$ $$\ce{NaOH}$$ solution (although it doesn't matter here). Actual definition is $$\pu{50 g}$$ of $$\ce{NaOH}$$ in $$\pu{100 g}$$ of solution. Since water is the solvent, your interpretation is correct.
Your second mistake is the lethal one. You have considered water and solid $$\ce{NaOH}$$ are additive. Assuming $$V_\text{total} = V_\ce{NaOH} + V_\ce{H2O}$$ is a mistake.
Using a given density of $$\ce{NaOH}$$ ($$\pu{1.5298 g}$$),we can calculate the volume of $$\pu{1000 g}$$ of solution: $$\frac{\pu{1000 g}}{\pu{1.5298 g/mL}} = \pu{653.68 mL}$$. Therefore, dissolving $$\pu{500 g}$$ of $$\ce{NaOH}$$ in $$\pu{500 mL}$$ of water (assuming temperature is $$\pu{25 ^\circ C}$$, hence density of water is $$\pu{1.00 g/mL}$$), the volume increased by only $$\pu{153.68 mL}$$. Do you still think volumes are additive?