Chester makes 1L of a 0.01M HCl solution and measure the pH. Burt then adds 5.80 g of sodium chloride
to Chesterโs solution and measures the pH. Calculate the difference between the two pH values
obtained.
This is not a simple question.
1. Use the Debye-Huckel relationship to calculate the pH of 0.01M HCl. That is not the simple pH = -log(0.01) = 2
2. Now do the same for the mixture of 0.01M HCl and 5.80g NaCl after converting NaCl to mols then to M.
3. Solve for delta pH.
To calculate the difference between the two pH values obtained, we first need to find the pH values of the two solutions.
1. Finding the pH of the initial HCl solution:
Chester prepares a 0.01M HCl solution with a volume of 1L. The concentration of the HCl solution is given as 0.01M.
The pH of a strong acid, such as HCl, can be calculated using the formula: pH = -log[H+]
Since HCl is a strong acid, it completely dissociates in water, so the concentration of HCl is equal to the concentration of H+ ions.
Thus, the pH of the HCl solution is: pH = -log(0.01) = -log10^-2 = 2
2. Finding the pH of the solution after adding sodium chloride:
Burt adds 5.80 g of sodium chloride to Chester's solution. Sodium chloride (NaCl) dissociates in water to form Na+ and Cl- ions. The Cl- ions react with water to produce HCl through hydrolysis.
The additional HCl formed due to the hydrolysis will affect the pH of the solution.
To calculate the new pH, we need to consider the concentration of the additional HCl formed.
To find the concentration of HCl formed, we need to calculate the moles of NaCl reacted and use the stoichiometry of the reaction.
The molar mass of NaCl is 22.99 + 35.45 = 58.44 g/mol.
The moles of NaCl can be calculated as follows: moles = mass / molar mass
moles of NaCl = 5.80 g / 58.44 g/mol = 0.099 mol
Since 1 mole of NaCl produces 1 mole of HCl, the concentration of the additional HCl formed is also 0.099 M.
Now, we can calculate the pH of the solution after adding sodium chloride using the new concentration of HCl:
pH = -log(0.01 + 0.099) = -log(0.109) = -log10^-0.96 = 0.96
Finally, to calculate the difference between the two pH values obtained, we subtract the pH of the initial solution from the pH of the solution after adding sodium chloride:
Difference in pH = (pH after adding NaCl) - (pH of initial HCl solution)
Difference in pH = 0.96 - 2 = -1.04
Therefore, the difference between the two pH values obtained is -1.04.