56.4k views
2 votes
What is the pH of a 2.1 M solution of HClO4?

User Borduhh
by
8.5k points

2 Answers

3 votes

Answer:

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

User Iembry
by
8.1k points
3 votes
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.
User Jason Jones
by
7.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.