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.
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
2 answers
How the concentration of HAc in first solution became 0.5M
Can you explain pls
Can you explain pls