Answer:
NH₃/NH₄Cl
Explanation:
We can calculate the pH of a buffer using the Henderson-Hasselbalch's equation.
![pH=pKa+log\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
If the concentration of the acid is equal to that of the base, the pH will be equal to the pKa of the buffer. The optimum range of work of pH is pKa ± 1.
Let's consider the following buffers and their pKa.
- CH₃COONa/CH3COOH (pKa = 4.74)
The optimum buffer is NH₃/NH₄Cl.
The answer is ................................ c
0.17 M is the is the molal concentration of this solution
Explanation:
Data given:
freezing point of glucose solution = -0.325 degree celsius
molal concentration of the solution =?
solution is of glucose=?
atomic mass of glucose = 180.01 grams/mole
freezing point of glucose = 146 degrees
freezing point of water = 0 degrees
Kf of glucose = 1.86 °C
ΔT = (freezing point of solvent) - (freezing point of solution)
ΔT = 0.325 degree celsius
molality =?
ΔT = Kfm
rearranging the equation:
m = 
m= 0.17 M
molal concentration of the glucose solution is 0.17 M
Answer:
Since in a chloride ion, we have an additional electron
you might think that it will affect the mass but the mass of an electron is almost negligible so we will ignore that
Amount of ions in 1 mol = 6.022 * 10^23
Amount of ions in 0.486 moles = 0.486 * (6.022*10^23)
Amunt of ions in 0.486 moles = 2.9 * 10^23 ions
Hence, option 1 is correct