Solution :
For the reaction :

we have
![$Ka = \frac{[\text{Tris}^- \times H_3O]}{\text{Tris}^+}$](https://tex.z-dn.net/?f=%24Ka%20%3D%20%5Cfrac%7B%5B%5Ctext%7BTris%7D%5E-%20%5Ctimes%20H_3O%5D%7D%7B%5Ctext%7BTris%7D%5E%2B%7D%24)


Clearing
, we have 
So to reach
, one must have the
concentration of the :
![$\text{[OH}^-]=10^{-pOH} = 6.31 \times 10^{-7} \text{ moles of base}$](https://tex.z-dn.net/?f=%24%5Ctext%7B%5BOH%7D%5E-%5D%3D10%5E%7B-pOH%7D%20%3D%206.31%20%5Ctimes%2010%5E%7B-7%7D%20%5Ctext%7B%20moles%20of%20base%7D%24)
So we can add enough of 1 M NaOH in order to neutralize the acid that is calculated above and also adding the calculated base.


Volume NaOH 
Tris mass 
Now to prepare the said solution we must mix:
gauge to 1000 mL with water.
Answer:
22.6According to the law of conservation of mass, the total mass of reactants should be equal to the total mass of products. Here, total mass of reactants= 11.3+11.3=22.6
•
• • Mass of Hcl produced=22.6 g
Answer:
1. hydrogen - H
2. helium - He
3. sodium - Na
4. magnesium - Mg
5. potassium - K
Explanation:
Hydrogen is the element of group 1 and first period. The atomic number of hydrogen is 1 and the symbol of the element is H.
The electronic configuration of the element hydrogen is:-

Helium is the element of group 18 and first period. The atomic number of helium is 2 and the symbol of the element is He.
The electronic configuration of the element helium is:-

Sodium is the element of group 1 and third period. The atomic number of sodium is 11 and the symbol of the element is Na.
The electronic configuration of the element sodium is:-

Magnesium is the element of group 2 and third period. The atomic number of magnesium is 12 and the symbol of the element is Mg.
The electronic configuration of the element magnesium is:-

Potassium is the element of group 1 and forth period. The atomic number of potassium is 19 and the symbol of the element is K.
The electronic configuration of the element potassium is:-

<span>We look at the end of the day:
n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
n(NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol
[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
M [OH-] = Kb * [NH3] = 0.070652*1.8*10^(-5) = 1.27174 x 10^(-6)
pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>