Strong acid-titrated with strong base. suppose the titration was reversed in question 2. if you 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? add this curve to your sketch in question 2. 3. la. 1.0; b. 7.0; c. 60 ml]

(a) pH = 1

(b) pH = 7

(c) 60 mL

Explanation:

(a) At the start of the titration, there is only 0.1 M HCl, which is a strong acid, according to the following expression.

HCl → H⁺ + Cl⁻

The concentration of H⁺ is 0.1 M as well and the pH is:

pH = -log [H⁺] = -log 0.1 = 1

(b) The reaction between HCl and NaOH is:

HCl + NaOH → NaCl + H₂O

At the equivalence point, all the HCl and NaOH have reacted. Since NaCl does not contribute to the pH, the pH is 7.

(c) The moles of HCl are:

(0.1 mol/L) × (30.0 × 10⁻³ L) = 3 × 10⁻³ mol

The molar ratio of HCl to NaOH is 1:1, so we also have 3 × 10⁻³ moles of NaOH. The volume of NaOH that contains these moles is:

3 × 10⁻³ mol / (0.1 mol/L) = 0.03 L = 30 mL

The total volume is:

V(HCl) + V(NaOH) = 30 mL + 30 mL = 60 mL

the answer is na (sodium)