Answer : The value of 
of the weak acid is, 4.72
Explanation :
First we have to calculate the moles of KOH.
Now we have to calculate the value of of the weak acid.
The equilibrium chemical reaction is:
Initial moles 0.25 0.03 0
At eqm. (0.25-0.03) 0.03 0.03
= 0.22
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[HK]}{[HA]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BHK%5D%7D%7B%5BHA%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the value of of the weak acid is, 4.72
Answer:
Because it does not have to
Explanation:
Why because when you blow at it it might move it does not change it's shape because it is a solid
Answer:
b. 2 mol of KI in 500. g of water
Explanation:
We have to apply the colligative property of freezing point depression.
The formula is: ΔT = Kf . m . i
As the (Kf . m . i) is higher, then the freezing temperature will be lower.
i refers to the Van't Hoff factor (number of ions dissolved in the solution)
KI → K⁺ + I⁻ (i =2)
Kf is constant so, we have to search for the highest m (molality)
Molality means the moles of solute in 1kg of solvent.
The highest m is option b → 2 mol of KI / 0.5 kg = 4 mol/kg
a. 1 mol of KI / 0.5 kg = 2 mol/kg
c. 1 mol of KI / 1kg = 1 mol/kg
d. 2 mol of KI / 1kg = 2 mol/kg
1000 g = 1kg. In order to determine molality we need to convert the mass (g) of solvent to kg
