Answer:
a) The pH of the solution is 12.13.
b) The pH of the solution is 12.17.
Explanation:
Ionic product of water =
![K_w=[H^+][OH^-]](https://tex.z-dn.net/?f=K_w%3D%5BH%5E%2B%5D%5BOH%5E-%5D)
![1.01\times 10^-{14}=[H^+][OH^-]](https://tex.z-dn.net/?f=1.01%5Ctimes%2010%5E-%7B14%7D%3D%5BH%5E%2B%5D%5BOH%5E-%5D)
Taking negative logarithm on both sides:
![-\log[1.01\times 10^-{14}]=(-\log [H^+])+(-\log [OH^-])](https://tex.z-dn.net/?f=-%5Clog%5B1.01%5Ctimes%2010%5E-%7B14%7D%5D%3D%28-%5Clog%20%5BH%5E%2B%5D%29%2B%28-%5Clog%20%5BOH%5E-%5D%29)
The pH is the negative logarithm of hydrogen ion concentration in solution.
The pOH is the negative logarithm of hydroxide ion concentration in solution.
a) 
of NaOH.
Concentration of hydroxide ions:
So, ![[OH^-]=1\times [NaOH]=1\times 1.39\times 10^{-2} M=1.39\times 10^{-2} M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1%5Ctimes%20%5BNaOH%5D%3D1%5Ctimes%201.39%5Ctimes%2010%5E%7B-2%7D%20M%3D1.39%5Ctimes%2010%5E%7B-2%7D%20M)
![pOH=-\log[1.39\times 10^{-2} M]=1.86](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5B1.39%5Ctimes%2010%5E%7B-2%7D%20M%5D%3D1.86)
pH=13.99-1.86=12.13
b) 
of NaOH.
Concentration of hydroxide ions:
So, ![[OH^-]=3\times [Al(OH)_3]=3\times 0.0051 M=0.0153 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D3%5Ctimes%20%5BAl%28OH%29_3%5D%3D3%5Ctimes%200.0051%20M%3D0.0153%20M)
![pOH=-\log[0.0153 M]=1.82](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5B0.0153%20M%5D%3D1.82)
pH=13.99-1.82=12.17
Answer:
3 to 5
Explanation:
The pH scale is from 0 to 14.
Acids have a pH of anything less than 7.
Anything with a pH greater than 7 is considered to be a base.
If a solution has a pH of 7, it is neutral.
3 to 5 is in the range of less than 7, so a solution of an acid can have this pH.
6 to 8 includes less than 7 , 7 and greater than 7, so it is not just in the range of an acid.
9 to 11 and 12 to 14 are both in the range of greater than 7, so they have to be bases.
Therefore, the correct answer is 3 to 5.
Answer:
V₂ = 530.5 mL
Explanation:
Given data:
Initial temperature = 20.0°C
Final temperature = 40.0 °C
Final volume = 585 mL
Initial volume = ?
Solution:
Initial temperature = 20.0°C (20+273 = 293 K)
Final temperature = 40.0 °C (40+273 = 323 K)
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₁ = V₂T₁ /T₂
V₂ = 585 mL × 293 K / 323 K
V₂ = 171405 mL.K / 323 K
V₂ = 530.5 mL
