How would the calculated concentration of the HCI be affected, if the sodium hydroxide were poured from a beaker that contained some water before that NaOH were added to it ?
Hmmmm. Wondering if the NaOH were solid, or in a solution?
hmmm. adding water dilutes the HCl
adding NaOH neutralizes the HCl
so, what do you think?
Seems like an MCAT question. You would over estimate the concentration of the acid.
To understand how the calculated concentration of HCl might be affected when sodium hydroxide (NaOH) is poured from a beaker that contains water before adding it, we need to consider the concept of dilution.
When you pour NaOH from a beaker that already contains water, you are essentially diluting the NaOH solution. Dilution is the process of adding a solvent (in this case, water) to a solution to decrease its concentration. The concentration of a solution is the amount of solute (in this case, NaOH) dissolved in a given amount of solvent (water).
Now, let's consider the reaction between NaOH and HCl:
NaOH + HCl → NaCl + H₂O
This balanced equation shows that 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl and 1 mole of water.
When you add diluted NaOH solution to HCl solution, the amount of water present in the NaOH will increase the total volume of the solution. As a result, the concentration of the HCl solution will decrease.
However, it's important to note that the final calculated concentration will depend on the initial concentrations of the solutions, the volumes of each solution used, and the balanced equation for the reaction. By applying the principles of stoichiometry and understanding the balanced equation, you can calculate the new concentration of HCl based on the amounts of NaOH and water added.
Therefore, if you pour NaOH from a beaker that contains water before adding it to HCl, the concentration of HCl will decrease due to the dilution effect caused by the additional water in the NaOH solution.