[JasonPOC]

To determine the concentration of solution of NaHCO3 in a titration against a standard solution HCl. the first step in this experiment is to accurately dilute 100.0 ml of 1.00 M of HCl stock solution to a 0.100M solution using a 1L volumetric flask. However, instead of using distilled water in the dilution , mistakenly add 900.0 mL of 0.0222 M Of NaOH solution

How to calculate the concentration of HCl in 1 L volumetric flask after adding NaOH solution ?
What is the principle of it ?
Thanks!

[Borek]

Trivial stoichiometry. What was the reaction that took place?

[JasonPOC]

Is it
NaOH + HCl ———> NaCl + H2O

[Borek]

Yes. Now it is a matter of finding what is left after the reaction.

[JasonPOC]

But H2O and NaCl will not affect the concentration of
As they are neutral findings

Can you use few steps to explain the question because I haven't done this type of  question before?
Thanks

[Borek]

Which is why you have to find out what was LEFT after the reaction.

Hint: have you heard about limiting reagents?

[JasonPOC]

Yup
I have heard it
In this case NaOH is a limiting reagent

The no. of mol of NaOH = the no. of mol of HCl
Calculate it and divide it with the volume of solution inthe flask
Then the answer is the concentration of HCl

[Borek]

In this case NaOH is a limiting reagent

Yes.

The no. of mol of NaOH = the no. of mol of HCl

That will allow you to calculate how many moles of HCl reacted (a necessary step), but you are interested in how many moles were left.