Answer:
pH= 9.2
Explanation:
Henderson hasselbach equation
pKa= log Ka= log (4.9 x 10^-10)=9.3
![pH=Pka+log \frac{[A-]}{[HA]}](https://tex.z-dn.net/?f=pH%3DPka%2Blog%20%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D)
![pH=9.3+log \frac{[CN-]}{[HCN]}](https://tex.z-dn.net/?f=pH%3D9.3%2Blog%20%5Cfrac%7B%5BCN-%5D%7D%7B%5BHCN%5D%7D)
![pH=9.3+log \frac{[0.64 M]}{[0.83 M]}](https://tex.z-dn.net/?f=pH%3D9.3%2Blog%20%5Cfrac%7B%5B0.64%20M%5D%7D%7B%5B0.83%20M%5D%7D)
pH= 9.2
<u>Answer:</u> No crystals of potassium sulfate will be seen at 0°C for the given amount.
<u>Explanation:</u>
We are given:
Mass of potassium nitrate = 47.6 g
Mass of potassium sulfate = 8.4 g
Mass of water = 130. g
Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g
This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water
Applying unitary method:
In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams
So, in 130 grams of water, the amount of potassium sulfate dissolved will be 
As, the soluble amount is greater than the given amount of potassium sulfate
This means that, all of potassium sulfate will be dissolved.
Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.
The pair that results in the most effective buffer is .50 M Ha and .50 M A-.
<span>A </span>buffer's<span> capacity
is the pH range where it works as an </span>effective buffer, preventing large changes in pH upon addition of
an acid or base.
The correct answer between all
the choices given is the second choice or letter B. I am hoping that this
answer has satisfied your query and it will be able to help you in your
endeavor, and if you would like, feel free to ask another question.
5g NaOH x 1 mol NaOH/ 39.99g NaOH = 0.125 mol NaOH