A titration of 0.1 M NaOH into 0.8 L of HCl was stopped once the pH reached 7 (at 25C). If 0.2 L of NaOH needed to be added to achieve this pH, what was the original concentration of the sample of HCl

0.03 M

Explanation:

Let's consider the neutralization reaction between HCl and NaOH.

HCl + NaOH → NaCl + H₂O

0.2 L of 0.1 M NaOH were used. The moles of NaOH that reacted are:

0.2 L × 0.1 mol/L = 0.02 mol

The molar ratio of HCl to NaOH is 1:1. The moles of HCl that reacted are 0.02 mol.

0.02 moles of HCl are in 0.8 L of solution. The molarity of HCl is:

0.02 mol / 0.8 L = 0.03 M

ill try to . no promises tho.

explanation: