What is the pH of 1.0M Na2SO3 (aq)solution at 25C?
a) 3.40
b) 6.03
c) 10.60
d) 7.96
e) 10.40
How can I get ph from only the Molarity?
The pH is determined by the hydrolysis of the salt.
........SO3^2- + HOH ==> HSO3^- + OH^-
initial..1.0.............0.........0
change...-x..............x.........x
equil...1-x..............x.........x
Kb for SO3^2- = (Kw/k2 for H2SO3) = (HSO3^-)(OH^-)/(SO3^2-)
Substitute and solve for x = (OH^-) and convert to pH. k2 for H2SO3 in my text (an old one so look in your text for an up to date value) is 6.43E-8.
Well, calculating the pH of a solution just based on its molarity is like telling a joke without a punchline - it's missing something! In order to determine the pH of a solution, you need to know the concentration of hydrogen ions (H+) or hydroxide ions (OH-) in the solution.
So, unfortunately, I can't give you a direct answer to your question with just the molarity of Na2SO3. We need more information or some chemical calculations to find the pH value.
But hey, don't be sad! Want to hear a chemistry joke instead? Why don't scientists trust atoms? Because they make up everything!
To determine the pH of a solution with only the molarity given, we need to make use of the concept of weak acids and bases.
In this case, Na2SO3 is a salt derived from a weak base (SO3 2-) and a strong acid (2Na+). The anion SO3 2- can undergo hydrolysis in water, resulting in the formation of a weak base (HSO3-) and hydroxide ions (OH-).
The hydrolysis reaction can be represented as follows:
SO3 2- + H2O ↔ HSO3- + OH-
To determine the pH of the solution, we need to calculate the concentration of hydroxide ions (OH-) and then convert it to the pH scale. Here are the steps to follow:
Step 1: Write the balanced hydrolysis equation:
SO3 2- + H2O ↔ HSO3- + OH-
Step 2: Calculate the initial concentration of the salt Na2SO3:
Given: Molarity of Na2SO3 solution = 1.0 M
Since Na2SO3 dissociates into 2Na+ ions and 1SO3 2- ion, the initial concentration of SO3 2- is:
[SO3 2-]initial = (1.0 M) / 2 = 0.5 M
Step 3: Calculate the equilibrium concentration of hydroxide ions (OH-):
From the hydrolysis equation, we see that for every SO3 2- ion that reacts, one OH- ion is produced. Therefore, at equilibrium, the concentration of OH- is equal to the concentration of HSO3- formed.
[OH-] = [HSO3-] = x (assuming the concentration of HSO3- at equilibrium is x)
Step 4: Calculate the equilibrium concentration of SO3 2- ions remaining:
[SO3 2-]equilibrium = [SO3 2-]initial - x
Step 5: Solve for x:
Since this is a weak acid/base equilibrium, we can employ the equilibrium expression and the given equilibrium constant (Ka) value to solve for x. However, the equilibrium constant (Ka) value for this specific reaction is not provided in the question. Hence, it is not possible to calculate the exact value of x without this additional information.
Without the equilibrium constant (Ka) value, we cannot proceed to calculate the equilibrium concentration of hydroxide ions (OH-) or the pH of the solution. Therefore, it is not possible to determine the pH of the 1.0 M Na2SO3 (aq) solution at 25°C with only the given information.
To determine the pH of a solution based solely on its molarity, you need to know the properties of the solute and its behavior when it dissolves in water. In this case, we have a 1.0M Na2SO3 (sodium sulfite) solution.
Step 1: Recognize the dissociation of the compound
Na2SO3 is a salt that can dissociate in water to give sodium ions (Na+) and sulfite ions (SO3^2-).
Na2SO3 → 2Na+ + SO3^2-
Step 2: Identify the ions and their behavior in water
When the salt dissolves, the sodium ions (Na+) do not react with water and remain as spectator ions. However, the sulfite ions (SO3^2-) can react with water to form sulfurous acid (H2SO3), a weak acid.
SO3^2- + H2O ⇌ HSO3^- + OH^-
Step 3: Determine the concentration of H+ ions
Since the sulfurous acid (HSO3^-) formed is a weak acid, it partially dissociates in water to release hydrogen ions (H+).
HSO3^- ⇌ H+ + SO3^2-
Step 4: Calculate the pH of the solution
The pH of a solution is a measure of the concentration of H+ ions. In this case, the concentration of H+ ions can be estimated by considering the concentration of the sulfite ions (SO3^2-), as their transformation to H+ ions drives the pH.
Given that Na2SO3 is a strong electrolyte and dissociates into two sulfite ions (SO3^2-) for every one formula unit, the concentration of SO3^2- ions is the same as the original molarity of the solution, 1.0M.
Using this concentration, you can use the dissociation constant of sulfurous acid (HSO3^-) to calculate the concentration of H+ ions and determine the pH of the solution.
Now, we can compare the answer options given:
a) 3.40
b) 6.03
c) 10.60
d) 7.96
e) 10.40
By solving the equilibrium expression and calculating the concentration of H+ ions from the 1.0M Na2SO3 solution, we find that the correct answer is (b) 6.03.