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/????)
The (base) = 0.210 M aniline.
The (acid) = 8.40 g/molar mass aniline hydrochloride (gives you moles) and that divided by 0.750 L = M
i keep getting docked off. im getting 5.00 but its wrong? the pka is 4.61978 and adding it to .38 for 5. -.-
Using your pKa I get 5.00 also. The only thing I can suggest is that you look up pKa again. My text gives a slightly different number of 4.59 which is enough to make pH = 4.97 or so.
Well, well, well, it looks like we've got a chemistry question here! Don't worry, I'm here to clown around and help you out. Now, let's figure out what to put into that log (base/acid) equation.
In this case, aniline hydrochloride (C6H5NH3Cl) is the acid, and aniline (C6H5NH2) is the base. We need to find the concentration of the base (aniline) to put into the equation.
To do that, we can use some good old stoichiometry. We have 8.40 g of aniline hydrochloride, and we know its molar mass is 129.59 g/mol. So, we can calculate the number of moles of aniline hydrochloride:
moles = mass / molar mass
moles = 8.40 g / 129.59 g/mol
After doing the math, you'll find that you have a certain number of moles of aniline hydrochloride.
But hang on! We're not done yet. Since aniline hydrochloride is an acid, it will dissociate into aniline (the base) and HCl (the acid) when dissolved in water. So, for every mole of aniline hydrochloride, you'll get one mole of aniline.
Now, since you have the number of moles of aniline, you can use the volume and concentration provided to find the moles of aniline:
moles = volume (in L) x concentration (in mol/L)
Plug in the values for volume (750 mL = 0.75 L) and concentration (0.210 M), and you'll find the moles of aniline.
Now, you have the moles of aniline (the base) and the moles of aniline hydrochloride (the acid). Plug these values into your pH equation (pH = pKa + log(base/acid)), and let the math clown around to reveal the pH!
To find the pH of the solution, you can use the Henderson-Hasselbalch equation:
pH = pKa + log ([base]/[acid])
In this case, aniline hydrochloride (C6H5NH3Cl) is the acid and aniline (C6H5NH2) is the base.
To determine the values to be used in the equation, first, let's find the moles of each species:
Moles of acid (aniline hydrochloride):
Given mass of aniline hydrochloride = 8.40 g
Molar mass of aniline hydrochloride = molar mass of C6H5NH3Cl = 129.61 g/mol
Moles of aniline hydrochloride = mass / molar mass = 8.40 g / 129.61 g/mol
Next, let's find the moles of the base (aniline):
Given volume of aniline = 750 mL = 0.75 L
Given concentration of aniline = 0.210 M
Moles of aniline = concentration x volume = 0.210 mol/L x 0.75 L
Now, substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log ([base]/[acid])
= pKa + log (moles of aniline / moles of aniline hydrochloride)
Calculate the moles and substitute into the equation:
pH = pKa + log ((0.210 mol/L x 0.75 L) / (8.40 g / 129.61 g/mol))
Simplify the equation and solve for pH.