What is the pH of the solution produced by mixing equal volume of 2*10^-3M HClO4 and 0.01M KClO4 ?

Question

Answer

Problem: What is the pH of the solution produced by mixing equal volume of 2*10^-3M HClO4 and 0.01M KClO4 ?

Solution:

The pH of the solution can be determined by using the following steps:

Step 1: Write the Balanced Chemical Equation

HClO4 + KClO4 → K+ + ClO4- + H+

Step 2: Calculate the Concentration of H+

The concentration of H+ can be calculated using the following equation:

[H+] = Ka × [HClO4] ÷ [HClO4] + [KClO4]

where Ka is the acid dissociation constant of HClO4, [HClO4] is the initial concentration of HClO4, and [KClO4] is the initial concentration of KClO4.

Using the given values:

Ka = 7.5 × 10^-8

[HClO4] = 2 × 10^-3 M

[KClO4] = 0.01 M

Substituting these values into the equation:

[H+] = (7.5 × 10^-8) × (2 × 10^-3) ÷ (2 × 10^-3 + 0.01)

[H+] = 9.6 × 10^-5 M

Step 3: Calculate the pH

The pH can be calculated using the following equation:

pH = -log[H+]

Substituting the value of [H+]:

pH = -log(9.6 × 10^-5)

pH = 4.02

Answer: The pH of the solution produced by mixing equal volume of 2*10^-3M HClO4 and 0.01M KClO4 is 4.02.

Answer

The answer is 2.7

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