What is the pH of a solution prepared by dissolving 1.23g of 2-nitrophenol (FM 139.11) in 0.250 L?
Well so, i calculated the no.moles of 2-nitrophenol, and divided it by 0.250, to get the molarity which was 3.53E-2.
Since the molar ratio of dissociation is 1:1, i assumed [H+] to be 3.53E-2.
and then using pH=-log[H+], i got pH to be 1.45
Can you please confirm if my answer or method is good? I'll be very grateful
Your solution assumes that 2-nitrophenol is a strong acid and dissociates completely (100%). But I found a pKa on the web listed as 7.2 which means Ka = 6.3E-8. If we call this compound HP, then
...........HP ==> H^+ + P^-
I.......0.0353....0......0
C.........-x......x.....x
E......0.0353-x...x......x
Then Ka = (H^+)(P^-)/(HP)
6.3E-8 = (x)(x)/(0.0353-x)
and solve for x = (H^+), then convert that to pH.
Your method is almost correct, but there are a couple of important things to consider.
First, when calculating the molarity, you need to divide the number of moles of solute by the volume of the solution in liters, not by the volume of the solvent. In this case, you dissolved 1.23 g of 2-nitrophenol in 0.250 L of solution, so you should use 0.250 L as the volume of the solution in the calculation.
Second, you correctly determined that the molar ratio of dissociation is 1:1, but you made an error in assuming that the concentration of hydrogen ions ([H+]) is equal to the molarity of the solution. This assumption is incorrect because not all of the solute will dissociate into hydrogen ions. You need to consider the dissociation constant (Ka) of 2-nitrophenol to calculate the actual concentration of hydrogen ions.
To calculate the concentration of hydrogen ions, you can use the expression for the dissociation constant (Ka):
Ka = [H+][A-] / [HA]
In this case, the concentration of hydrogen ions ([H+]) is equal to the concentration of the dissociated form of 2-nitrophenol, which is A-. Assuming the concentration of undissociated 2-nitrophenol (HA) is equal to the molarity of the solution (3.53E-2 M), you can rearrange the equation to solve for [H+] as follows:
[H+] = (Ka * [HA]) / [A-]
You can then use the equation pH = -log[H+] to calculate the pH of the solution.
So, the correct method would be to:
1. Convert the mass of 2-nitrophenol to moles using its molar mass (FM).
2. Divide the moles of solute by the volume of the solution (0.250 L) to get the molarity.
3. Use the dissociation constant (Ka) of 2-nitrophenol to calculate the concentration of hydrogen ions ([H+]).
4. Finally, use the equation pH = -log[H+] to determine the pH of the solution.
I hope this clarifies the process for you!
To confirm if your answer and method are correct, let's go through the steps together:
Step 1: Calculate the number of moles of 2-nitrophenol.
Given the mass of 2-nitrophenol is 1.23g and its formula mass is 139.11 g/mol, we can calculate the number of moles using the formula:
moles = mass / molar mass
moles = 1.23g / 139.11 g/mol
moles ≈ 0.00884 mol
Step 2: Calculate the molarity of the solution.
The molarity (M) is defined as moles of solute divided by the volume of the solution in liters. Since the volume is given as 0.250 L, we can calculate the molarity using the formula:
Molarity = moles / volume
Molarity = 0.00884 mol / 0.250 L
Molarity ≈ 0.0354 M
Based on your calculation, you obtained a molarity of 3.53E-2, which is the same as 0.0353 M. So your calculation for the molarity is correct.
Step 3: Determine the concentration of H+ ions.
Since the molar ratio of the dissociation is 1:1, the concentration of H+ ions is equal to the molarity of the solution.
Based on your assumption, [H+] = 0.0353 M.
Step 4: Calculate the pH.
The pH is calculated by taking the negative logarithm (base 10) of the concentration of H+ ions.
pH = -log[H+]
pH = -log(0.0353)
pH ≈ 1.45
Therefore, your answer of pH ≈ 1.45 is correct.
In conclusion, your method and answer are both correct. Well done!