# Calculate the pH and concentration of H_3O of a 0.285 M HCl solution.

## Question:

Calculate the pH and concentration of H{eq}_3 {/eq}O of a 0.285 M HCl solution.

## pH of a Solution:

pH of a solution gives us an indication of the concentration of the hydronium ion {eq}H_3O^{+} {/eq} in the solution. This helps us to determine the nature of the solution (acidic/basic/neutral). The value is measured using a generalized scale known as the pH indicator ranging from 0 to 14.

It is calculated by taking the negative logarithm of the {eq}H_3O^{+} {/eq} ion concentration in the solution.

Thus we have {eq}pH = - log \left [ H_3O^{+} \right ] {/eq}

## Answer and Explanation:

Given:

Concentration of {eq}HCl {/eq} is {eq}\displaystyle 0.285 \ M {/eq}.

Now, we write the dissociation reaction of the solution in water

{eq}HCl +H_2O \ (aqueous)\rightleftharpoons H_3O^++Cl ^- {/eq}

We know that hydrochloric acid {eq}HCl {/eq} is a very strong acid.

Here we see that the concentration of {eq}H_3O^+ {/eq} ion is the same as the concentration of the {eq}HCl {/eq}

{eq}[H_3O^+] = 0.285 \ M {/eq}.

{eq}\begin{align*} pH& = - log [H_3O^+] \\ & = -log \ (0.285 ) \\ & = 0.545 \Rightarrow(Answer) \end{align*} {/eq}

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from Physical Science: Middle School

Chapter 4 / Lesson 4