Question

What is the pH of a 2.1 M solution of HClO4?

Answer 1

3.61

Step-by-step explanation:

`K_(a)=2.9* 10^(-8)`

Concentration = 2.1 M

Consider the ICE take for the dissociation of acetic acid as:

HClO ⇄ H⁺ + ClO⁻

At t=0 2.1 - -

At t =equilibrium (2.1-x) x x

The expression for dissociation constant of acetic acid is:

`K_(a)=\frac {\left [ H^(+) \right ]\left [ {ClO}^- \right ]}{[HClO]}`

`2.9* 10^(-8)=\frac {x^2}{2.1-x}`

x is very small, so (2.1 - x) ≅ 2.1

Solving for x, we get:

x = 2.47×10⁻⁴ M

pH = -log[H⁺] = -log(2.47×10⁻⁴) = 3.61

Answer 2

Since HCl04 is a strong acid, being [H+], and a molarity of 2.1 M.

To solve for pH:

pH = -log (M) = -log(2.1M) = -0.32, which is clearly a negative number.

However, to verify the answer, just use the pH meter in determining the pH of the solution.