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
Answer:
1.79 mol.
Explanation:
- For the balanced reaction:
<em>2NaCl + F₂ → 2NaF + Cl₂.
</em>
It is clear that 2 mol of NaCl react with 1 mol of F₂ to produce 2 mol of NaF and 1 mol of Cl₂.
- Firstly, we can get the no. of moles of F₂ gas using the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 1.2 atm).
V is the volume of the gas in L (V = 18.3 L).
n is the no. of moles of the gas in mol (n = ??? mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (299 K).
∴ no. of moles of F₂ (n) = PV/RT = (1.2 atm)(18.3 L)/(0.0821 L.atm/mol.K)(299 K) = 0.895 mol.
- Now, we can find the no. of moles of NaCl is needed to react with 0.895 mol of F₂:
<em><u>Using cross multiplication:</u></em>
2 mol of NaCl is needed to react with → 1 mol of F₂, from stichiometry.
??? mol of NaCl is needed to react with → 0.895 mol of F₂.
∴ The no. of moles of NaCl needed = (2 mol)(0.895 mol)/(1 mol) = 1.79 mol.
To answer your question, the answer is number 1. hope this help