Substance X (MW=356gm) is a weak acid. Its Ka is 1.7x10^-3. Calculate pH of725ml of the following solutions:
a)Containing 0.187 mol of substance x
b)containing 127gm of Substance x
I have been trying to figure this problem out for hours now and do not get it. Please help me and be very detailed. Thank you.
Call the weak acid HA.
(HA) = mols/L = 0.187/0.725 = about 0.26M but you should be more accurate than that.
.......HA --> H^+ + A^-
I.....0.26M....0.....0
C.....-x.......x.....x
E....-0.26-x...x.....x
Ka = 1.7E-3 = (H^+)(A^-)/(HA)
Substitute into Ka expression and solve for x = (H^+) then convert to pH.
Part B is done the same way EXCEPT you don't have mols to start. You convert 127g to mols by mols = grams/molar mass, then M = mols/L solution and from there do the same process as above.
Post your work if you get stuck.
When I solve for H^+ I put 1.7 x10^-3 x 0.26. I got 4.42 x 10^-4. My pH is 3.35. Is that right?
To solve this problem, we need to use the expression for the dissociation of a weak acid in water and apply the Henderson-Hasselbalch equation. Let's go step-by-step through each part of the question to find the pH.
a) To calculate the pH of a solution containing 0.187 mol of Substance X, we first need to determine the concentration of Substance X in the solution.
Given:
- Volume of the solution = 725 mL = 0.725 L
- Moles of Substance X = 0.187 mol
To find the concentration (C) of Substance X in moles per liter (mol/L) or molarity, we divide the moles by the volume:
C = moles/volume
C = 0.187 mol / 0.725 L
C ≈ 0.258 mol/L
Now, let's use the Henderson-Hasselbalch equation to find the pH. The Henderson-Hasselbalch equation is given as:
pH = pKa + log([A-]/[HA])
Where:
- pH is the measure of acidity or basicity of a solution
- pKa is the negative logarithm of the acid dissociation constant (Ka)
- [A-] is the concentration of the conjugate base
- [HA] is the concentration of the weak acid
For Substance X, the weak acid can be represented as HA, and its conjugate base as A-. The equation for the dissociation of Substance X in water (under acidic conditions) is:
HA ⇌ H+ + A-
Given:
- Ka (acid dissociation constant) = 1.7 x 10^-3
Now, substitute the known values into the Henderson-Hasselbalch equation:
pH = -log(Ka) + log([A-]/[HA])
We need to calculate the [A-]/[HA] ratio. Since Substance X is a weak acid, it will be mostly undissociated, and we can approximate the ratio as [A-]/[HA] ≈ 0.
Therefore, the equation simplifies to:
pH = -log(Ka)
Substituting the value of Ka:
pH = -log(1.7 x 10^-3)
pH ≈ -(-2.77)
pH ≈ 2.77
Therefore, the pH of a solution containing 0.187 mol of Substance X is approximately 2.77.
b) To calculate the pH of a solution containing 127 g of Substance X, we need to convert the mass of Substance X to moles using its molar mass (MW), which is given as 356 g/mol.
Given:
- Mass of Substance X = 127 g
- Molar mass of Substance X = 356 g/mol
To find the moles of Substance X, we divide the mass by the molar mass:
moles = mass/MW
moles = 127 g / 356 g/mol
moles ≈ 0.356 mol
Now, let's follow the same steps as in part (a) to find the pH.
First, calculate the concentration (C) of Substance X:
C = moles/volume
C = 0.356 mol / 0.725 L
C ≈ 0.490 mol/L
Next, use the Henderson-Hasselbalch equation:
pH = -log(Ka) + log([A-]/[HA])
Since [A-]/[HA] ≈ 0 for weak acids, the equation simplifies to:
pH = -log(Ka)
Substituting the value of Ka:
pH = -log(1.7 x 10^-3)
pH ≈ -(-2.77)
pH ≈ 2.77
Therefore, the pH of a solution containing 127 g of Substance X is approximately 2.77.