Answer : The concentration of
and
are
and
respectively.
Solution : Given,
pH = 4.10
pH : pH is defined as the negative logarithm of hydronium ion concentration.
Formula used : ![pH=-log[H_3O^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%2B%5D)
First we have to calculate the hydronium ion concentration by using pH formula.
![4.10=-log[H_3O^+]](https://tex.z-dn.net/?f=4.10%3D-log%5BH_3O%5E%2B%5D)
![[H_3O^+]=antilog(-4.10)](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3Dantilog%28-4.10%29)
![[H_3O^+]=7.94\times 10^{-5}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D7.94%5Ctimes%2010%5E%7B-5%7D)
Now we have to calculate the pOH.
As we know, 


Now we have to calculate the hydroxide ion concentration.
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
![9.9=-log[OH^-]](https://tex.z-dn.net/?f=9.9%3D-log%5BOH%5E-%5D)
![[OH^-]=antilog(-9.9)](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3Dantilog%28-9.9%29)
![[OH^-]=1.258\times 10^{-10}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.258%5Ctimes%2010%5E%7B-10%7D)
Therefore, the concentration of
and
are
and
respectively.
We will assume that the solvent is water. So, if we have 100 grams of the solution, 19 grams will be sodium hydroxide, while the remaining 81 grams will be water.
The molar weight of sodium hydroxide, NaOH, is 40. The molar weight of water is 18. Finding the moles of each:
NaOH:
19 / 40 = 0.475
Water:
81 / 18 = 4.5
Total moles present:
4.5 + 0.475 = 4.975 moles
The mole fraction of NaOH is:
0.475 / 4.975 = 0.0955
The mole fraction of NaOH is 0.0955
Answer:
Celsius +273.15
Explanation:
The two scales have the same size degree, but Kelvin is an absolute scale based on absolute zero while 0° in Celsius is based on the melting point of water. So, in order to convert from Celsius degrees to Kelvin we only need to add 273.15 to the given temperature:
K= °C + 273.15
Answer:
0.014 theoretical yield and 79% percentage yield
Answer:
new solubility = 0.443 g/l
Explanation:
To solve this question, we will use Henry's law.
Henry's law states that, at constant temperature:
S1 / P1 = S2 / P2
We are given that:
S1 = 0.16 g/l
P1 = 104 kPa
S2 is the solubility we want to calculate
P2 = 288 kPa
Substitute with the givens in the above relation to get the new solubility (S2) as follows:
0.16 / 104 = S2 / 288
S2 = (0.16/104) * 288
S2 = 0.443 g/l
Hope this helps :)