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

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.