HCl is titrated with NaOH. When doing the titration, some of the NaOH splashed onto the inside surface of the Erlenmyer flask, and you forgot to rinse it into your sample. Would the systematic error be falsely high, low, or unaffected and why? Thanks.

As I understand the problem, you have HCl in the Erlenmeyer flask and you are titrating with NaOH from a buret. So if you don't wash all of the NaOH into the HCl, you will be adding more NaOH from the buret to make up for that which stayed on the walls of the flask. That makes the NaOH volume too high. How would that affect the molarity of the HCl?
vol HCl x M HCl = vol NaOH x M NaOH.
M HCl = (vol NaOH x M NaOH)/vol HCl

In this case, if you don't wash all of the NaOH into the HCl and some NaOH remains on the inside surface of the Erlenmeyer flask, it will lead to a falsely high molarity of HCl.

To understand why, let's consider the equation for the titration:

(vol HCl x M HCl) = (vol NaOH x M NaOH)

The molarity of HCl (M HCl) is equal to the ratio of the volume of NaOH (vol NaOH) multiplied by its molarity (M NaOH) to the volume of HCl (vol HCl).

Now, if some of the NaOH splashes onto the inside surface of the flask and is not rinsed into the sample, the volume of NaOH used in the calculation will be greater than the actual volume of NaOH required to neutralize the HCl.

This means that the calculated molarity of HCl will be higher than its actual value. Therefore, the systematic error in this case would be a falsely high molarity of HCl.