A student dissolved the KHP in 100.0mL of water instead of 50.0mL. Will the molar concentration of the NaOH solution determined from titrating be greater, less than, or unaffected by this mistake? Explain.

To determine whether the molar concentration of the NaOH solution will be greater, less than, or unaffected by the mistake, we need to understand the relationship between the amount of KHP dissolved and the molar concentration of the NaOH solution.

In a titration, a known volume and concentration of one solution (in this case, NaOH) is reacted with an unknown volume and concentration of another solution (in this case, KHP). The reaction between NaOH and KHP follows a 1:1 stoichiometric ratio.

To calculate the molar concentration of the NaOH solution, we use the equation:

Molarity of NaOH = (moles of KHP) / (volume of NaOH solution used in the titration)

Now, the mistake in this scenario is that the student dissolved the KHP in 100.0mL of water instead of 50.0mL. Since the molar concentration of a solution is defined as the amount of solute (in moles) divided by the volume of the solution (in liters), the change in volume affects the concentration.

Specifically, when the student dissolved the KHP in 100.0mL of water instead of the intended 50.0mL, the final volume of the KHP solution used in the titration is greater than what was initially planned.

As a result, the molar concentration of the NaOH solution will be affected. It will be **less** than what it should be if the student had dissolved the KHP in the correct volume (i.e., 50.0mL). This is because the molar concentration of the NaOH solution is inversely proportional to the volume of the KHP solution used in the titration.

In summary, the mistake of dissolving the KHP in 100.0mL of water instead of 50.0mL will result in a lower molar concentration of the NaOH solution compared to the intended value.