Respuesta :

MissPhiladelphia
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.

Answer:

3.61

Explanation:

[tex]K_{a}=2.9\times 10^{-8}[/tex]

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:

[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {ClO}^- \right ]}{[HClO]}[/tex]

[tex]2.9\times 10^{-8}=\frac {x^2}{2.1-x}[/tex]

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

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