Calculate the pH of a 0.250M solution of Sulfurous acid, H2SO3.
You will need the Ka for sulfurous acid to calculate the pH as it is a weak acid. Were you given a figure in the question?
H2SO3<> HSO3- + H+
Ka = 1.54×10−2 mol L^-1
(from a data book)
at start
H2SO3 HSO3- H+
0.250M 0M 0M
at equilibrium
0.250-x x x
Ka=[HSO3-][H+]/[H2SO3]
1.54×10−2 mol L^-1 = x^2/(0.250-x)
you can either solve the quadratic or assume that x is small with respect to 0.250 hence
1.54×10−2 mol L^-1 = x^2/(0.250 mol L^-1)
x^2=0.00385 mol^2 L^-2
hence find x
and pH=-log(x)
1.207
To calculate the pH of a solution of sulfurous acid (H2SO3), we can use the equilibrium expression for its ionization in water.
The chemical equation representing the ionization of sulfurous acid is:
H2SO3(aq) ⇌ H+(aq) + HSO3-(aq)
From this equation, we can determine that one hydrogen ion (H+) is released for every molecule of sulfurous acid that ionizes.
The equilibrium expression for this ionization can be written as:
Ka = [H+(aq)][HSO3-(aq)] / [H2SO3(aq)]
Given that the initial concentration of H2SO3 is 0.250 M, we can assume that the concentration of H+ and HSO3- formed is x (M) at equilibrium.
Plugging in these values into the equilibrium expression, we get:
Ka = x * x / 0.250
The Ka value for sulfurous acid is 1.7 x 10^-2. Rearranging the equation, we have:
x^2 = Ka * 0.250
x = sqrt(Ka * 0.250)
Substituting the Ka value, we get:
x = sqrt(1.7 x 10^-2 * 0.250)
x ≈ 0.086 M
Since we assumed that the concentration of H+ and HSO3- formed is x (M), we can approximate the concentration of H+ as 0.086 M.
The pH is defined as the negative logarithm (base 10) of the concentration of hydrogen ions. Hence,
pH = -log[H+]
pH = -log(0.086)
pH ≈ 1.07
Therefore, the pH of a 0.250 M solution of sulfurous acid is approximately 1.07.
To calculate the pH of a solution of Sulfurous acid, H2SO3, you need to consider the acidic dissociation of the acid in water and the equilibrium constant associated with this dissociation.
The dissociation of Sulfurous acid in water can be represented by the following equation:
H2SO3 + H2O ⇌ H3O+ + HSO3-
The equilibrium constant for this reaction is known as the acid dissociation constant, Ka, which is given by:
Ka = [H3O+][HSO3-] / [H2SO3]
First, you need to determine the concentration of the H3O+ ion in the solution. Since Sulfurous acid is a weak acid, we can assume that the concentration of H3O+ at equilibrium is equal to the concentration of H2SO3 that dissociates. Let's denote this concentration as x.
Initially, the concentration of the undissociated acid, H2SO3, is 0.250 M. Thus, at equilibrium, the concentration of H2SO3 becomes 0.250 - x M, and the concentration of the HSO3- ion also becomes x M.
Now, substitute the concentrations into the equation for the Ka value:
Ka = [H3O+][HSO3-] / [H2SO3]
Ka = x * x / (0.250 - x)
Since the value of Ka for Sulfurous acid is known to be 1.7 x 10^(-2), we can rearrange the equation and set it up as a quadratic equation:
1.7 x 10^(-2) = x^2 / (0.250 - x)
Simplifying this equation, we get:
x^2 = (1.7 x 10^(-2)) * (0.250 - x)
x^2 = 4.25 x 10^(-3) - 1.7 x 10^(-2) x
Rearrange the equation:
x^2 + 1.7 x 10^(-2) x - 4.25 x 10^(-3) = 0
Now, we can solve this quadratic equation either by factorizing or by using the quadratic formula. In this case, it is easier to use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Where:
a = 1
b = 1.7 x 10^(-2)
c = -4.25 x 10^(-3)
Plugging in the values:
x = (-1.7 x 10^(-2) ± √((1.7 x 10^(-2))^2 - 4(1)(-4.25 x 10^(-3)))) / (2(1))
Simplifying further, we obtain two possible values for x:
x₁ ≈ 0.064 M (ignoring the negative root, as it is not physically meaningful)
Therefore, the concentration of H3O+ in the solution at equilibrium is approximately 0.064 M.
Finally, calculate the pH using the equation -log[H3O+]:
pH = -log(0.064)
pH ≈ 1.19
Therefore, the pH of a 0.250 M solution of Sulfurous acid, H2SO3, is approximately 1.19.