Find the ratio of the pH of solution (I) containing 1 mol of CH3COONa and 1 mol of HCl in 1L and solution (II)
containing 1 mol of CH3COONa and 1 mol of CH3COOH in 1L
In equal quantities like this I would write it as
NaAc + HCl ==> HAc + NaCl
So you essentially have a solution of HAc (now not 1M but 0.5M) and you calculate the pH of that solution. Do the same for solution II using the HH equation, then find the ratio of the pH o the two solutions.
How the concentration of HAc in first solution became 0.5M
Can you explain pls
To find the ratio of the pH of solution (I) containing 1 mol of CH3COONa and 1 mol of HCl in 1L, and solution (II) containing 1 mol of CH3COONa and 1 mol of CH3COOH in 1L, we need to understand the acid-base properties of these compounds.
CH3COONa is the sodium salt of acetic acid (CH3COOH), and it is a strong electrolyte that completely dissociates into its ions in water (Na+ and CH3COO-). HCl is a strong acid that also completely dissociates into H+ and Cl- ions in water.
In solution (I), the reaction between CH3COONa and HCl can be written as follows:
CH3COONa + HCl -> NaCl + CH3COOH
In solution (II), the reaction between CH3COONa and CH3COOH can be written as follows:
CH3COONa + CH3COOH -> CH3COONa + CH3COOH
In both cases, the CH3COONa remains unchanged while forming CH3COOH.
Now, let's analyze the pH of each solution:
1. Solution (I) containing 1 mol of CH3COONa and 1 mol of HCl:
Since both CH3COONa and HCl are strong electrolytes, they will completely dissociate in water. HCl will contribute H+ ions to the solution, while CH3COONa will not contribute any extra H+ ions. Therefore, the resulting solution will have a high concentration of H+ ions, making it highly acidic. The pH of this solution will be low.
2. Solution (II) containing 1 mol of CH3COONa and 1 mol of CH3COOH:
Here, CH3COOH is a weak acid that does not fully dissociate in water. CH3COONa, being a salt of CH3COOH, will provide CH3COO- ions. The CH3COOH and CH3COO- ions will establish an equilibrium in the solution, which dictates the acidity of the solution. The pH will be determined by the dissociation constant, Ka, of CH3COOH.
To calculate the pH in solution (II), we need the value of Ka, or the dissociation constant, for CH3COOH.
Once we have the value of Ka for acetic acid (CH3COOH), we can use the equilibrium equation and the concentrations of CH3COOH and CH3COO- to calculate the pH of the solution. The specific calculations can be done using the Henderson-Hasselbalch equation or a more detailed equilibrium expression.
Therefore, to find the ratio of the pH between the two solutions, we need to obtain the Ka value for acetic acid and then perform the necessary calculations.
Keep in mind that the ratio of the pH values of the solutions will depend on the specific values of the concentrations and the Ka of acetic acid.