What is the ph at each of the points in the titration of 25.00 ml of 0.2000 m
hcl by 0.2000 m naoh:
i) before adding naoh
ii) after adding 24.00 ml naoh

i) pH = 0.6990

ii) pH = 2.389

Explanation:

i) Before adding aqueous NaOH, there are 25.00 mL of 0.2000 M HCl. HCl reacts with the water in the aqueous solution as follows:

HCl + H₂O ⇒ H₃O⁺ + Cl⁻

The HCl and H₃O⁺ are related to each other through a 1:1 molar ratio, so the concentration of H₃O⁺ is equal to the HCl concentration.

The pH is related to the hydronium ion concentration as follows:

pH = -log([H₃O⁺]) = -log(0.2000) = 0.699

ii) Addition of NaOH causes the following reaction:

H₃O⁺ + NaOH ⇒ 2H₂O + Na⁺

The H₃O⁺ and NaOH react in a 1:1 molar ratio. The amount of NaOH added is calculated:

n = CV = (0.2000 mol/L)(24.00 mL) = 4.800 mmol NaOH

Thus, 4.800 mmol of H₃O⁺ were neutralized.

The initial amount of H₃O⁺ present was:

n = CV = (0.2000 mol/L)(25.00 mL) = 5.000 mmol H₃O⁺

The amount of H₃O⁺ that remains after addition of NaOH is:

(5.000 mmol) - (4.800 mmol) = 0.2000 mmol

The concentration of H₃O⁺ is the amount of H₃O⁺ divided by the total volume. The total volume is (25.00 mL) + (24.00 mL ) = 49.00 mL

C = n/V = (0.2000 mmol) / (49.00 mL) = 0.004082 M

The pH is finally calculated:

pH = -log([H₃O⁺]) = -log(0.004082) = 2.389

lol idk man i guess u gotta ask bill nye the science guy