No, There are not likely to form in areas that experience a lot of erosion on a daily basis.
We are given the molar concentration of an aqueous solution of weak acid and the pH ofthe solution, and we are asked to determine the value of Ka for the acid.
The first step in solving any equilibrium problem is to write the equation for the equilibriumreaction. The ionization of benzoic acid can be written as seen in the attached image (1).
The equilibrium-constant expression is the equation number (2)
From the measured pH, we can calculate pH as seen in equation (3)
To determine the concentrations of the species involved in the equilibrium, we imagine that thesolution is initially 0.10 M in HCOOH molecules. We then consider the ionization of the acidinto H+ and HCOO-. For each HCOOH molecule that ionizes, one H+ ion and one HCOO- ionare produced in solution. Because the pH measurement indicates that [H+] = 1x 10^-4 M atequilibrium, we can construct the following table as seen in the equation number (4)
To find the value of Ka, please see equation (5):
We can now insert the equilibrium concentrations into the expression for Ka as seen in equation (6)
Therefore, 2.58x10^-4 M is the concentration of benzoic acid to have a pH of 4.0
Answer : The molar mass of solute is, 89.9 g/mol
Explanation : Given,
Mass of solute = 5.8 g
Mass of solvent (water) = 100 g
Formula used :

where,
= change in freezing point
= temperature of pure solvent (water) = 
= temperature of solution = 
= freezing point constant of water = 
m = molality
Now put all the given values in this formula, we get


Therefore, the molar mass of solute is, 89.9 g/mol
Answer:
Mass = 199.21 g
Explanation:
Given data:
Moles of HCl = 3.59 mol
Mass of CaCl₂ = ?
Solution:
Chemical equation:
2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
we will compare the moles of HCl with CaCl₂ from balanced chemical equation:
HCl : CaCl₂
2 : 1
3.59 : 1/2×3.59 = 1.795
3.59 moles of HCl will produced 1.795 moles of CaCl₂.
Mass of CaCl₂.
Mass = number of moles × molar mass
Mass = 1.795 mol × 110.98 g/mol
Mass = 199.21 g