Explanation:
The given reaction is as follows.
Hence, number of moles of NaOH are as follows.
n = 
= 0.005 mol
After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.
n = 
= 0.0025 mol
According to ICE table,
Initial: 0.005 mol 0.0025 mol 0 0
Change: -0.0025 mol -0.0025 mol +0.0025 mol
Equibm: 0.0025 mol 0 0.0025 mol
Hence, concentrations of HA and NaA are calculated as follows.
[HA] = 
[NaA] =
![[A^{-}] = [NaA] = \frac{0.0025 mol}{V}](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%5D%20%3D%20%5BNaA%5D%20%3D%20%5Cfrac%7B0.0025%20mol%7D%7BV%7D)
Now, we will calculate the 
value as follows.
pH = 
![pK_{a} = pH - log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%3D%20pH%20-%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
= 
= 3.42
Thus, we can conclude that of the weak acid is 3.42.
add all the number and find the average then subtract the mass defect and then you will get your answer
The Relative Formula Mass of NaH2PO4 is 120 g/mol
Therefore, the number of moles = 6.6/120
= 0.055 moles of NaH2PO4 which is also equal to the number of moles of H2PO4.
[H2PO4-] = Number of moles oof H2PO4-/Volume of the solution in L
= 0.055/ ( 355 ×10^-3)
= 0.155 M
Na2HPO4 undergoes complete dissociation as follows;
Na2HPO4 (aq)= 2Na+ (aq) + HPO4^2- (aq)
1 mole of Na2HPO4 = 142 g/mol
Therefore; number of moles = 8.0/142
= 0.0563 moles
[HPO4 ^-2] is given by no of moles HPO4^2- /volume of the solution in L
= 0.0563/(355×10^-3)
= 0.1586 M
Both H2PO4^2- and HPO4^2- are weak acids the undergoes partial dissociation
Ka of H2PO4- = 6.20 × 10^-8
[H+] =Ka*([H2PO4-]/[HPO4(2-)]
= (6.20 ×10^-8)×(0.155/0.1586)
= 6.059 ×10^-8 M
pH = - log[H+]
= - log (6.059×10^-8)
= 7.218
The molality of a solute is equal to the moles of solute per kg of solvent. We are given the mole fraction of I₂ in CH₂Cl₂ is <em>X</em> = 0.115. If we can an arbitrary sample of 1 mole of solution, we will have:
0.115 mol I₂
1 - 0.115 = 0.885 mol CH₂Cl₂
We need moles of solute, which we have, and must convert our moles of solvent to kg:
0.885 mol x 84.93 g/mol = 75.2 g CH₂Cl₂ x 1 kg/1000g = 0.0752 kg CH₂Cl₂
We can now calculate the molality:
m = 0.115 mol I₂/0.0752 kg CH₂Cl₂
m = 1.53 mol I₂/kg CH₂Cl₂
The molality of the iodine solution is 1.53.
