Step-by-step explanation:
A buffer is a mixture of a weak acid and its conjugate base, or a weak base and its conjugate acid. In our case we are reacting a strong acid (HCl) and a weak base (sodium acetate). So we don't have a buffer. The strong acid will neutralize the weak base. We have to determine which of them is in excess and find the pH of the resulting solution.
HCl (aq) + CH₃COONa (aq) ----> NaCl (aq) + CH₃COOH (aq)
First we have to determine the number of moles of each reactant. We added 275 ml of a 0.676 M solution of HCl and 500 mL of a 0.525 M solution of CH₃COONa. We can use the definition of molarity concentration to determine the number of moles of each reagent.
Molarity = moles of solute/volume of solution in L
moles of solute = molarity * volume of solution in L
moles of HCl = 0.676 M * 0.275 L
moles of HCl = 0.186 moles
moles of CH₃COONa = 0.525 M * 0.500 L
moles of CH₃COONa = 0.262 moles
HCl (aq) + CH₃COONa (aq) ----> NaCl (aq) + CH₃COOH (aq)
In the equation of the reaction all the coefficients are 1. So 1 mol of HCl will completely neutralize 1 mol of CH₃COONa. The molar ratio between them is 1 to 1.
1 mol of HCl = 1 mol of CH₃COONa
We mixed 0.262 moles of CH₃COONa with 0.186 moles of HCl. The 0.186 moles of HCl will neutralize 0.186 moles of CH₃COONa. And CH₃COONa will be excess.
Excess of CH₃COONa = 0.262 moles - 0.186 moles
Excess of CH₃COONa = 0.076 moles
Now we have to determine the concentration of this excess. We mixed 0.275 L with 0.500 L. Then the total volume of solution is 0.775 L. And the concentration of CH₃COONa after the reaction is:
total volume = 0.500 L + 0.275 L
total volume = 0.775 L
Resulting molarity of CH₃COONa = 0.076 moles/0.775 L
Resulting molarity of CH₃COONa = 0.098 M
Finally to get our answer we have to determine the pH of this resulting solution. To determine the pH of a weak base we have to use the ICE table. In solution
Answer: