Question

A 25.0 mL sample of 0.723 M HClO4 is titrated with a 0.273 M KOH solution. What is the [H ] (molarity) before any base is added

Answer 1

0.723 is the [H ] (molarity) before any base is added.

the balanced equation for the acid-base reaction is

KOH + HClO₄ ---> KClO₄ + H₂O

stoichiometry of KOH to HClO₄ is 1:1

The number of HClO₄ moles - 0.723 M / 1000 mL/L x 25.0 mL = 0.0181 mol

Number of KOH moles - 0.27 M/ 1000 mL/L x 80.0 mL = 0.0216 mol

KOH is a strong acid and HClO₄ is a strong base therefore complete dissociation takes place.

the acid reacts with base in a 1:1 molar ratio, there's excess base remaining.

excess OH⁻ ions - 0.0216 - 0.0181 = 0.0035 mol

concentration is calculated as the number of moles/volume

volume of solution - 25.0 + 80.0 = 105.0 mL

[OH⁻] = 0.0035 mol / 0.105 L = 0.033 M

pOH = -log [OH⁻]

pOH = -log(0.033 M)

pOH = 1.48

pH can be calculated by knowing the pOH

pH + pOH = 14

pH = 14 - 1.48 = 12.52

pH = -log [H₃O⁺]

[H₃O⁺] = antilog(-12.52)

[H₃O⁺]  = 3.0 x 10⁻¹³ M

0.723 is the [H ] (molarity) before any base is added.

To know more about molarity refer to:  brainly.com/question/8732513

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