Answer:
substance A is a liquid, while substance B is a liquid or gas
Answer:
the symbol that is missing might be 2.
Explanation:
I am not 100% on this, so correct me if I am wrong.
Answer:
The dissociation constant of phenol from given information is
.
Explanation:
The measured pH of the solution = 5.153

Initially c
At eq'm c-x x x
The expression of dissociation constant is given as:
![K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BC_6H_5O%5E-%5D%5BH%5E%2B%5D%7D%7B%5BC_6H_5OOH%5D%7D)
Concentration of phenoxide ions and hydrogen ions are equal to x.
![pH=-\log[x]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5Bx%5D)
![5.153=-\log[x]](https://tex.z-dn.net/?f=5.153%3D-%5Clog%5Bx%5D)



The dissociation constant of phenol from given information is
.
The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 1.25 mL
<u>Explanation:</u>
To calculate the volume of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of stomach acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

Hence, the volume of HCl neutralized is 1.25 mL