What is the pH of a 100.0 mL 0.325 M H3BO3 solution at 25 C? (Why is this an acidic solution?)
H3BO3 is a weak acid because it ionizes to provide H^+ in solution. Ka1 = 7.2E-10. Ka2 is 1.8E-13 but for pH of this solution need not be considered since it is 1000 times weaker than Ka1 and contributes very little on its own.
................H3BO3 ==> H^+ + [H2BO3]^-
I..............0.325............0..............0
C................-x...............x...............x
E.............0.325-x..........x...............x
Plug the E line into the Ka expression and solve for x = (H^+), then convert this to pH by pH = -log*H^+).
Post your work if you get stuck.
To find the pH of a solution, we need to know the concentration of hydronium ions (\(H_3O^+\)) present in the solution. In this case, we are given the concentration of the acid H3BO3.
The first step is to write the balanced chemical equation for the ionization of H3BO3:
\[H3BO3 + H2O \rightleftharpoons H3O^+ + B(OH)_3\]
H3BO3 can be considered a weak acid, so it partially ionizes in water. The hydronium ion \((H_3O^+)\) is responsible for the acidity of the solution.
Given that the initial concentration of H3BO3 is 0.325 M, we can assume that the concentration of \(H_3O^+\) will be the same since it is a 1:1 ratio between the H3BO3 and \(H_3O^+\) ions.
Since pH is defined as the negative logarithm (base 10) of the concentration of hydronium ions, we can use the equation:
\[pH = -\log [H_3O^+]\]
Substituting the concentration of \(H_3O^+\), we get:
\[pH = -\log (0.325)\]
Using a calculator, we find that the pH of the solution is approximately 0.49.
This solution is considered acidic because the pH is below 7. A pH of 0.49 indicates a relatively high concentration of hydronium ions, which corresponds to an acidic solution.
To find the pH of the H3BO3 solution, we first need to understand the properties of H3BO3 and how it dissociates in water.
H3BO3, also known as boric acid, is a weak acid. When it dissolves in water, it partially dissociates into ions:
H3BO3 ⇌ H+ + H2BO3-
In this dissociation, one proton (H+) is released, making H3BO3 an acid. The resulting H+ ion is what gives the solution its acidic nature.
To find the pH of the solution, we can use the equation:
pH = -log[H+]
where [H+] represents the concentration of hydrogen ions in moles per liter.
To calculate the concentration of H+ ions, we need to know the dissociation constant of H3BO3 and the equilibrium concentrations of the acid and its conjugate base (H2BO3-). The K a expression for the dissociation of H3BO3 can be written as:
[H+][H2BO3-] / [H3BO3] = K a
The acid dissociation constant (K a) for boric acid is 5.8 × 10^-10 mol/L.
Now, let's solve for the equilibrium concentrations:
We have a 0.325 M H3BO3 solution.
Given that H3BO3 is a weak acid, we can assume that x (the change in concentration) is negligible compared to the initial concentration of H3BO3. As a result, we can set up the following equilibrium expression:
[x][x] / [0.325] ≈ 5.8 × 10^-10
Since we're neglecting x compared to 0.325, the concentration of H2BO3- can be considered roughly equal to x. Substituting this approximation, we get:
(x^2) / 0.325 ≈ 5.8 × 10^-10
Simplifying further, we solve for x:
x^2 ≈ 0.325 × (5.8 × 10^-10)
x^2 ≈ 1.885 × 10^-10
x ≈ √(1.885 × 10^-10)
x ≈ 4.344 × 10^-6 mol/L
So, the concentration of [H+] ions in the solution is approximately 4.344 × 10^-6 mol/L.
Now, let's find the pH using the equation pH = -log[H+]:
pH = -log(4.344 × 10^-6)
pH ≈ 5.363
Therefore, the pH of the 0.325 M H3BO3 solution is approximately 5.363.
The solution is considered acidic because the pH value is less than 7. In general, any solution with a pH below 7 is considered acidic. In this case, boric acid partially ionizes, releasing H+ ions into the solution, which increases the concentration of H+ and lowers the pH, thus making it acidic.