What is the pH of a 2.1 M solution of HClO4?
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:
\(K_{a}=2.9\times 10^{-8}\)
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:
\(K_{a}=\frac {\left [ H^{+} \right ]\left [ {ClO}^- \right ]}{[HClO]}\)
\(2.9\times 10^{-8}=\frac {x^2}{2.1-x}\)
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