Question

What is the pH of 250.0mL solution that is 0.0955 M HCH3CO2 (Ka=1.8x10-5) and 0.125 M NaCH3CO2?

a. Now, 20.0 mL of 0.455 M HCl (aq) is added to the original buffer.

i. Write the neutralization reaction (use the actual species in the reaction).

ii. Find the moles of all species. iii. Build a stoichiometry table with the reaction and the data (or anything that shows what happens in the reaction).

iv. What is the pH of the solution? Was your prediction correct?

v. Do you still have a buffer? Why?

Answer 1

pH= pKa + log[CH3CO2-]/[HCH3CO2]

pKa= -logKa= 4.74

pH= 4.74 + log 0.125/0.0955 = 4.86

a) i) HCl + NaCH3CO2 -----> HCH3CO2 + NaCl

ii) I don´t undertand if the question ask for moles before or after the addition of HCl, i will calculate before because in part iii) asks for a table showing what happens.

mol HCl= 0.455M x 0.020L= 9.1x10^-3mol

mol NaCH3CO2= 0.125M x 0.25L= 0.03125 mol

mol HCH3CO2= 0.0955 x 0.25L= 0.023875 mol

iii) The HCl will react with the conjugated base of HCH3CO2 in order to produce acid.

HCl + NaCH3CO2 -----> HCH3CO2 + NaCl

9.1x10^-3mol__0.03125 mol__0.023875 mol__0

0___ 0.03125- 9.1x10^-3mol__0.023875+9.1x10^-3mol __9.1x10^-3mol

iv) final mol HCH3CO2= 0.023875+9.1x10^-3mol= 0.032975 mol --> Concentration= 0.032975 mol/0.270L= 0.122M

final mol CH3CO2- = 0.03125- 9.1x10^-3mol = 0.02215mol --> Concentration= 0.02215mol/0.270L= 0.082M

pH= 4.74 + log 0.082/0.122= 4.57

v) yes, the buffer capacity is lost when the addition of a strong acid or base change the pH in 1 unit, in this case the pH changed just 0.17 units.