Answer: pH=12.69
Explanation:



Initial 0.12 0 0
Eqm 0.12-x x x
![K_a=\frac{[H^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
(neglecting small value of x in comparison to 0.12)

Moles of 



0.06 moles of NaOH will give 0.06 moles of ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
Now
moles of
will be neutralized by
moles of
and
moles of
will be left.
Molarity of 
![pOH=-\log[OH^-]=-\log[0.049]=1.31](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D%3D-%5Clog%5B0.049%5D%3D1.31)
pH = 14 - pOH= 14 - 1.31 = 12.69
We can rephrase the statement with a little more specificity in order to understand the answer here.
The mass of the products can never be more than the The mass that is expected.
Answer:
The pH of the solution is 8.
Explanation:
To which options are correct, let us determine the concentration of the hydroxide ion, [OH-] and the pH of the solution. This is illustrated below:
1. The concentration of the hydroxide ion, [OH-] can be obtained as follow:
pOH = –Log [OH-]
pOH = 6
6 = –Log [OH-]
–6 = Log [OH-]
[OH-] = Antilog (–6)
[OH-] = 1x10^–6 mol/L
2. The pH of the solution can be obtained as follow:
pH + pOH = 14
pOH = 6
pH + 6 = 14
pH = 14 – 6
pH = 8.
From the calculations made above,
[OH-] = 1x10^–6 mol/L
pH = 8.
Therefore, the correct answer is:
The pH of the solution is 8
The graph is needed to answer this question.
Solubility may increase or decrease with temperature depending on the properties of the solute and the solvent.
It is quite common that the solubility of the ionic compounds, like KBr, in water increases with temperature.
Use your solubility curve for the KBr and you wiil see a line that starts at a solubility a little greater than 50 grams of the salt in 100 grams of water for temperaute 0°C and increase linearly until almost 100 grams of the salt in 100 grams of water at 100°C.
So, in this case you can affirm that the solubility of KBr increases with the temperature.
Answer: the second option: the solubility increases.