What is the pH of a solution in which 296 mL of HCl measured at 28.0 C and 1.02 atm, is dissolved in 1.5 L of aqueous solution?
You don't say but I assume the 296 mL HCl is a GAS. Then use PV = nRT and solve for n. Then M HCl = moles HCl/1.5L. Finally, pH = -log(H^+) and M HCl = H^+.
I plugged in all my numbers but I am still stuck on how to find n:moles
296mL is the volume
To determine the pH of the solution, we first need to calculate the concentration of HCl in the final solution.
To do this, we will use the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Given:
- Volume of HCl gas (V): 296 mL = 0.296 L
- Pressure (P): 1.02 atm
- Temperature (T): 28 °C = 28 + 273.15 = 301.15 K (convert to Kelvin)
Next, we need to calculate the number of moles (n) of HCl:
n = PV / RT
Substituting the values:
n = (1.02 atm * 0.296 L) / (0.0821 L·atm/mol·K * 301.15 K)
n ≈ 0.0128 moles
Now, we can calculate the concentration of HCl in the final solution:
Concentration (C) = moles / volume
C = 0.0128 mol / 1.5 L
C ≈ 0.00853 M
The concentration of HCl in the solution is approximately 0.00853 M.
Finally, we can use the pH scale to find the pH of the solution. The pH scale ranges from 0 to 14, with values below 7 being acidic. pH is calculated using the formula:
pH = -log[H+]
[H+] in this case refers to the concentration of the hydrogen ion (H+), which is the same as the concentration of HCl.
Therefore, the pH of the solution is:
pH = -log(0.00853)
pH ≈ 2.07
So, the pH of the solution is approximately 2.07.