Questions 36 and 37 refer to the following: during a laboratory activity, a student places 25.0 ml of hcioaq) of unknown concentration into a flask. the student adds four drops of phenolphthalein to the solution in the flask. the solution is titrated with 0.150 m koh(aq) until the solution appears faint pink. the volume of koh(aq) added is 18.5 ml. 36) complete the equation below for the neutralization reaction that occurs during the given titration. koh (aq) hcl(aq)- 37o determine the concentration of the hci(aq) solution, using the given titration data.

36)  HCl(aq)+KOH(aq)⇒H₂O(l)+KCl(aq)

37)  You can use the equation M₁V₁=M₂V₂
M₁=concentration of acid
V₁=volume of acid in orginal solution
M₂=concentration of base
V₂= the volume of based used to neutralize the acid
M₁=(0.150M KOH x 0.0185L KOH)/0.025L HCl
M₁=0.111M HCl
This equation works since we no based off of the formula in question 36 that 1 mole of KOH will neutralize 1 mole of HCl and M₂V₂ is equal to the moles KOH which means it needs to equal M₁V₁ since that is the number of moles HCl. (if moles HCl did not equal moles KOH the solution would not be neutralized).

Tolerance to high levels of salinity