Answer:
I'm sure it's option A. Cast Fossil
Answer : The pH of the solution is, 3.41
Explanation :
First we have to calculate the moles of
.


Now we have to calculate the value of
.
The expression used for the calculation of
is,

Now put the value of
in this expression, we get:



The reaction will be:

Initial moles 0.375 0.100 0.375
At eqm. (0.375-0.100) 0 (0.375+0.100)
= 0.275 = 0.475
Now we have to calculate the pH of solution.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[F^-]}{[HF]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
Now put all the given values in this expression, we get:
![pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]](https://tex.z-dn.net/?f=pH%3D3.17%2B%5Clog%20%5B%5Cfrac%7B%28%5Cfrac%7B0.475%7D%7B1.50%7D%29%7D%7B%28%5Cfrac%7B0.275%7D%7B1.50%7D%29%7D%5D)

Thus, the pH of the solution is, 3.41
Calorie (cal) or small calorie, is the amount of energy needed to heat one gram of water by one degree Celsius.
One small calorie is approximately 4.2 joules.
A calorie is a unit of energy.
For example:
Q = 150000 J; released energy.
Convert Jules in calories:
Q = 150000 J ÷ 4.186 J/calorie.
Q = 35833.73 calorie.
The large calorie or nutritional calorie (symbols: Cal, kcal) is the heat energy needed to raise the temperature of one kilogramof water by one degree Celsius.
A nutritional calorie, or kilocalorie, is equal to 1000 calories.
One large calorie is 4.184 kilojoules: 1 Cal = 4.184 kJ.
For example:
E = 2500 Cal; energy burnt.
E = 2500 Cal · 4.184 kJ/Cal.
E = 10460 kJ.