Answer:
A) pH of Buffer solution = 4.59
B) pH after 5.0 ml of 2.0 M NaOH have been added to 400 ml of the original buffer solution = 4.65
Explanation:
This is the Henderson-Hasselbalch Equation:
![pH = pKa + log\frac{[conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D)
to calculate the pH of the following Buffer solutions.
C, because a homogeneous mixture is evenly mixed.
All those others would be heterogeneous (not evenly mixed).
The full chemical symbol for an element<span> shows its mass </span>number<span> at the top, and its atomic </span><span>number at the bottom</span>
Answer:
Case 1:
X = Any element from Group I
i) H
ii) Li
iii) Na
iv) K
v) Rb
vi) Cs
Y = 1
Case 2:
X = Any element from Group II
i) Be
ii) Mg
iii) Ca
iv) Sr
v) Ba
vi) Ra
Y = 2
Case 3:
X = Any element from Group III
i) B
ii) Al
iii) Ga
iv) In
v) Ti
Y = 3
Explanation:
The general formula given is as follow,
XCly
So, if X has +1 oxidation state, then it will require only one Cl atom with oxidation number -1 to form a neutral compound, therefore, y = 1.
If X has +2 oxidation state, then it will require two Cl atoms with oxidation number -1 to form a neutral compound, therefore, y = 2.
If X has +3 oxidation state, then it will require three Cl atoms with oxidation number -1 to form a neutral compound, therefore, y = 3.
