Answer:
Explanation:
HA + H₂O ⇌ H₃O⁺ + A⁻
1. Calculate pKₐ
2. Calculate the pH
We can use the Henderson-Hasselbalch equation to get the pH.
![\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 4.89 +\log \left(\dfrac{0.42}{0.23}\right )\\\\& = & 4.89 + \log1.83 \\& = & 4.89 +0.261\\& = & 5.15\\\end{array}\\\text{The pH of the buffer is $\large \boxed{\mathbf{5.15}}$}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Ctext%7BpH%7D%20%26%20%3D%20%26%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%20%5Cleft%28%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5Cright%20%29%5C%5C%5C%5C%26%20%3D%20%26%204.89%20%2B%5Clog%20%5Cleft%28%5Cdfrac%7B0.42%7D%7B0.23%7D%5Cright%20%29%5C%5C%5C%5C%26%20%3D%20%26%204.89%20%2B%20%5Clog1.83%20%5C%5C%26%20%3D%20%26%204.89%20%2B0.261%5C%5C%26%20%3D%20%26%205.15%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20pH%20of%20the%20buffer%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B5.15%7D%7D%24%7D)
Ti + 2 Cl2 → TiCl4
(3.00 g Ti) / (47.867 g Ti/mol) = 0.062674 mol Ti
(6.00 g Cl2) / (70.9064 g Cl2/mol) = 0.084619 mol Cl2
0.084619 mole of Cl2 would react completely with 0.084619 x (1/2) = 0.0423095 mole of Ti, but there is more Ti present than that, so Ti is in excess and Cl2 is the limiting reactant.
(0.084619 mol Cl2) x (1 mol TiCl4 / 2 mol Cl2) x (189.679 g TiCl4/mol) = 8.025 g TiCl4 in theory
(7.7 g) / (8.025 g) = 0.96 = 96% yield TiCl4
Jupiter, Saturn, Uranus, Neptune, Jovian... is your answer.
