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.

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