Question

1. will the concentration of naoh solution increase or decrease if the solution is exposed to air for prolonged periods? briefly explain.

Explanation:

NaOH (sodium hydroxide) is a base. Air contains $\ce{CO_2}$ (carbon dioxide). When $\ce{NaOH}$ solution is exposed to air, $\ce{NaOH}$ reacts with $\ce{CO_2}$: $\ce{2NaOH + CO_2 = Na_2CO_3 + H_2O}$. This reaction consumes the $\ce{NaOH}$ solute. Concentration is calculated as $\frac{\text{moles of solute}}{\text{volume of solution}}$. As the amount of $\ce{NaOH}$ (solute) decreases while the volume of the solution (assuming negligible evaporation for the context of this reaction - dominated change) remains relatively constant (or changes minimally compared to the solute change), the concentration of the $\ce{NaOH}$ solution will decrease. Also, if evaporation occurs, but the reaction - based consumption of $\ce{NaOH}$ is more significant in terms of concentration change (since the reaction directly reduces the amount of the solute we are interested in, $\ce{NaOH}$), the net effect is a decrease in $\ce{NaOH}$ concentration.

Answer:

The concentration of the $\ce{NaOH}$ solution will decrease. This is because $\ce{NaOH}$ in the solution reacts with $\ce{CO_2}$ from the air (\(\ce{2NaOH + CO_2 = Na_2CO_3 + H_2O}\)), which consumes the $\ce{NaOH}$ solute. Since concentration is \(\frac{\text{moles of solute}}{\text{volume of solution}}\), a decrease in the moles of $\ce{NaOH}$ (solute) (with the solution volume remaining relatively constant) leads to a decrease in the concentration of the $\ce{NaOH}$ solution.