What is the pH of a solution prepared by dissolving 8.40 g of aniline hydrochloride C6H5NH3Cl in 750 mL of 0.210 M aniline,C6H5NH2?
i kno to use the ph = pka + log (base/acid). getting pka isnt the issue, but what do i put into the log (base/acid)? (0.210/????)
thanks
To find the pH of the solution, we need to determine the concentration of the base (aniline, C6H5NH2) and the acid (aniline hydrochloride, C6H5NH3Cl) in the solution.
First, let's calculate the concentration of the base, aniline (C6H5NH2), in the solution.
We are given that the volume of the solution is 750 mL and the concentration of aniline is 0.210 M.
To find the moles of aniline, we can use the equation:
moles = concentration × volume
moles of aniline = 0.210 M × 0.750 L = 0.1575 moles
Next, let's find the concentration of the acid, aniline hydrochloride (C6H5NH3Cl), in the solution.
We are given that the mass of aniline hydrochloride is 8.40 g. To calculate the moles of aniline hydrochloride, we need to know its molar mass. The molar mass of C6H5NH3Cl is:
6(12.01 g/mol) + 5(1.01 g/mol) + 1(14.01 g/mol) + 1(35.45 g/mol) = 131.62 g/mol
Using the molar mass, we can now calculate the moles of aniline hydrochloride:
moles of aniline hydrochloride = mass / molar mass = 8.40 g / 131.62 g/mol ≈ 0.0638 moles
Since aniline hydrochloride dissociates to form one mole of aniline (C6H5NH2) and one mole of HCl (hydrochloric acid, HCl), we have an equal concentration of acid (HCl).
Therefore, we have 0.0638 moles of acid (HCl) and 0.1575 moles of base (aniline).
Now we can write the equation:
pH = pKa + log10 (base/acid)
In this case, pKa is the value you obtained before.
So, the equation becomes:
pH = pKa + log10 (0.1575/0.0638)
To get the final answer, substitute the value of pKa you obtained and then calculate the logarithm using a calculator.
Note: Make sure to use logarithm base 10 (log10) for this calculation since it is the common logarithm used in chemistry.