A vitamin C tablet containing 250 mg of ascorbic acid (C6H7O6; Ka = 8.0 x 10-5 is dissolved in a 250 mL glass of water. What is the pH of the solution?
i tried answering this question multiple times but i still get it wrong.... can someone help? step-by-step please.
what i did was....
make a chemical equation
C6H8O6 + H20 <--> C6H7O6- + H30+
but i fell like my equation is off....
but i did an ICE table and used it to find x then the pH
HC is ascorbic acid with Ka = 8.0E-5.
moles HC = 0.250g/molar mass HC = ?.
M = moles/0.250L = ? which I will call y.
............HC ==> H^+ + C^-
initial.....yM......0.....0
change.......-x......x....x
equil.......y-x......x....x
Ka = (H^+)(C^-)/(HC)
Substitute and solve for x = (H^+) and convert to pH.
This looks like a low molarity; you MAY need to solve the quadratic.
Well, it seems like your equation is close, but it needs a small correction. Since ascorbic acid (C6H8O6) is a weak acid, it doesn't fully ionize in water. So, you can write the equation as follows:
C6H8O6 + H2O ⇌ C6H7O6- + H3O+
Now, let's create the ICE table:
Initial: C6H8O6 + H2O ⇌ C6H7O6- + H3O+
Change: -x -x +x +x
Equilibrium: C6H8O6 - x H2O - x ⇌ C6H7O6- + x H3O+ + x
The Ka expression for this equilibrium is:
Ka = [C6H7O6-][H3O+] / [C6H8O6]
The initial concentration of C6H8O6 is 0.250 g / (176.12 g/mol) = 0.00142 mol/L.
Since one molecule of C6H8O6 produces one C6H7O6- and one H3O+, the initial concentration of both is zero.
At equilibrium, the concentration of C6H8O6 will be (0.00142 - x) mol/L, and the concentrations of C6H7O6- and H3O+ will both be x mol/L.
Now, substitute these values into the Ka expression:
8.0 x 10^-5 = (x)(x) / (0.00142 - x)
Since x is small compared to 0.00142, you can assume that x is negligible compared to the initial concentration:
8.0 x 10^-5 = (x)(x) / (0.00142)
Simplifying this equation will give you a quadratic equation that you can solve to find the value of x. Once you find x, you can calculate the concentration of H3O+ and use that to determine the pH using the equation pH = -log[H3O+].
Remember, this is just a step-by-step guide to help you set up the problem. If you're still having trouble, feel free to ask for more assistance!
To find the pH of the solution, we need to consider the dissociation of ascorbic acid (C6H7O6) in water. The chemical equation for its dissociation is:
C6H7O6 + H2O ⇌ C6H7O6- + H3O+
Let's proceed step-by-step to find the pH.
Step 1: Write the balanced chemical equation for the dissociation of ascorbic acid.
C6H7O6 + H2O ⇌ C6H7O6- + H3O+
Step 2: Define the initial concentration of ascorbic acid.
The problem states that there is a 250 mg tablet of ascorbic acid dissolved in 250 mL of water. Since the molar mass of ascorbic acid is 176.13 g/mol, we can convert the mass to moles:
250 mg = 0.250 g
0.250 g / 176.13 g/mol = 0.00142 mol
Therefore, the initial concentration (C0) of ascorbic acid is:
C0 = 0.00142 mol / 0.250 L = 0.00568 M
Step 3: Define the change and equilibrium concentrations.
Using an ICE (Initial, Change, Equilibrium) table, we can determine the change and equilibrium concentrations. Since the initial concentration of ascorbic acid is larger than the equilibrium concentrations, we can assume that the change in concentration is negligible compared to the initial concentration. Therefore, we can assume that the equilibrium concentrations are:
[C6H7O6] = [H3O+] = x
[C6H7O6-] = x
Step 4: Write the equilibrium expression (Ka) for the dissociation reaction.
Ka = [C6H7O6-][H3O+] / [C6H7O6]
Given that Ka = 8.0 × 10^-5, we can substitute the equilibrium concentrations:
8.0 × 10^-5 = (x)(x) / (0.00568 - x)
Step 5: Solve the quadratic equation.
Rearranging the equation and assuming that x is small compared to 0.00568, we can approximate:
8.0 × 10^-5 ≈ (x^2) / (0.00568 - x)
Simplifying further:
8.0 × 10^-5 ≈ x^2 / 0.00568
8.0 × 10^-5 × 0.00568 ≈ x^2
0.0004544 ≈ x^2
Taking the square root of both sides:
sqrt(0.0004544) ≈ x
0.0213 ≈ x
Since x represents the concentration of H3O+, which contributes to pH, we have:
[H3O+] = 0.0213 M
Step 6: Calculate the pH.
The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration ([H3O+]):
pH = -log[H3O+]
Substituting the value of [H3O+]:
pH = -log(0.0213)
Using a scientific calculator, we find:
pH ≈ 1.67
Therefore, the pH of the solution is approximately 1.67.
To solve this problem step-by-step, we need to consider the dissociation of ascorbic acid (C6H7O6) in water. Here's how you can determine the pH of the solution:
Step 1: Write the chemical equation for the dissociation of ascorbic acid in water:
C6H7O6 + H2O ⇌ C6H7O6- + H3O+
Step 2: Since ascorbic acid acts as a weak acid, it does not fully ionize in water. We need to set up an ICE table to determine the concentration of the hydronium ion (H3O+).
Initial:
C6H7O6: 0.25 mol (calculated from the 250 mg of ascorbic acid)
H2O: Ignored (as its concentration remains constant)
Change:
C6H7O6: -x (as it reacts to form C6H7O6-)
H2O: Ignored
Equilibrium:
C6H7O6: 0.25 - x
C6H7O6-: x
H3O+: x
Step 3: Use the dissociation constant expression (Ka) to set up an equation:
Ka = [C6H7O6-][H3O+]/[C6H7O6]
From the equation above, we can substitute the equilibrium concentrations into the equation:
Ka = x*x/(0.25 - x)
Step 4: Since the value of x is assumed to be very small, we can neglect it in comparison to 0.25:
Ka = x*x/0.25
Step 5: Rearrange the equation and solve for x:
x^2 = Ka * 0.25
x = √(Ka * 0.25)
Step 6: Calculate the concentration of hydronium ions (H3O+):
[H3O+] = x
Step 7: Calculate the pH of the solution:
pH = -log[H3O+]
Keep in mind that the value of Ka you provided (8.0 x 10^-5) seems to be the equilibrium constant at a given temperature. You may need to consider the temperature in order to correctly calculate the pH.
I hope this step-by-step explanation helps you solve the problem.