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
The ability to attract an electron for bonding is called (option B) Electronegativity.
Answer:
1+2 = 12
Explanation:
this is a math equation, not a chemical formula
Answer:
The calculated concentration of sodium thiosulphate solution will be less than the actual value.
Explanation:
When IO3^2- solution is added to KI solution, I2 gas is released ,then sulphuric acid is now added to facilitate reduction. In order to prevent the escape of iodine (I2) gas ,the solution must immediately be titrated with thiosulphate.
If the solution is not immediately titrated with thiosulphate, the concentration of iodine available in the system decreases. When this occurs, it will also cause a decrease in the amount of iodine available to react with thiosulphate thus decreasing the concentration of thiosulphate obtained from calculation
