Answer:
A is correct option you can go for it
The question is incomplete, here is the complete question:
Calculate the pH at of a 0.10 M solution of anilinium chloride
. Note that aniline
is a weak base with a
of 4.87. Round your answer to 1 decimal place.
<u>Answer:</u> The pH of the solution is 5.1
<u>Explanation:</u>
Anilinium chloride is the salt formed by the combination of a weak base (aniline) and a strong acid (HCl).
To calculate the pH of the solution, we use the equation:
![pH=7-\frac{1}{2}[pK_b+\log C]](https://tex.z-dn.net/?f=pH%3D7-%5Cfrac%7B1%7D%7B2%7D%5BpK_b%2B%5Clog%20C%5D)
where,
= negative logarithm of weak base which is aniline = 4.87
C = concentration of the salt = 0.10 M
Putting values in above equation, we get:
![pH=7-\frac{1}{2}[4.87+\log (0.10)]\\\\pH=5.06=5.1](https://tex.z-dn.net/?f=pH%3D7-%5Cfrac%7B1%7D%7B2%7D%5B4.87%2B%5Clog%20%280.10%29%5D%5C%5C%5C%5CpH%3D5.06%3D5.1)
Hence, the pH of the solution is 5.1
Give me the question and I’ll see if I can help
Unfortunately the data provided doesn't include the DENSITY of the ammonium chloride solution and molarity is defined as moles per volume. So without the density, the calculation of the molarity is impossible. But fortunately, there are tables available that do provide the required density and for a 20% solution by weight, the density of the solution is 1.057 g/ml.
So 1 liter of solution will mass 1057 grams and the mass of ammonium chloride will be 0.2 * 1057 g = 211.4 g. The number of moles will then be 211.4 g / 53.5 g/mol = 3.951401869 mol. Rounding to 3 significant digits gives a molarity of 3.95.
Now assuming that your teacher wants you to assume that the solution masses 1.00 g/ml, then the mass of ammonium chloride will only be 200g, and that is only (200/53.5) = 3.74 moles.
So in conclusion, the expected answer is 3.74 M, although the correct answer using missing information is 3.95 M.