Strong acid-titrated with strong base. suppose the titration was reversed in question 2. if your titrated 30.0 ml of 0.1 m hcl with 0.1 m naoh, indicate the approximate ph (a) at the start of the titration and (b) at the equivalence point. (c) what is the total volume of solution at the equivalence point

(a) pH = 1.0

(b) pH = 7.0

(c) 60 mL

Explanation:

(a) At the start of the titration there's no OH⁻ species yet.

pH = -log[H⁺]

And because HCl is a strong acid

[H⁺] = [HCl] = 0.1 M

Thus

pH = -log (0.1) = 1.0

(b) In every strong base-strong acid titration, the pH at the equivalence point is 7.0.

(c) Because the concentration of HCl and NaOH are the same, you would need 30 mL of the NaOH solution to neutralize 30 mL of the HCl solution, thus the total volume is 60 mL.

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.