Gen Chem Question Buffers

[Miley13]

Advertisement - Members don't see this ad

This is a question I came across in Cliffs AP Chemistry Practice Exams and I was wondering if anyone could explain it. Thanks in advance!!


A buffer is formed by adding 500 mL of 0.20 M HC2H3O2 to 500 mL of 0.10 M NaC2H3O2. What would be the max amount of HCl that could be added to this solution without exceeding the capacity of the buffer?

a. 0.01 mol
b. 0.05 mol
c. 0.10 mol
d. 0.15 mol
e. 0.20 mol

I understand that C2H3O2- the weak base would have 0.05 moles found by multiplying (0.50 L) x (0.10 M) to get 0.05 moles. I know that acid and base react in a 1:1 ratio in buffers.

However, I am confused as to how the answer is B. I thought when you add a strong acid (HCl) to a weak base (C2H3O2-) you add half as much.

 

[lutopie0730]

This is a question I came across in Cliffs AP Chemistry Practice Exams and I was wondering if anyone could explain it. Thanks in advance!!


A buffer is formed by adding 500 mL of 0.20 M HC2H3O2 to 500 mL of 0.10 M NaC2H3O2. What would be the max amount of HCl that could be added to this solution without exceeding the capacity of the buffer?

a. 0.01 mol
b. 0.05 mol
c. 0.10 mol
d. 0.15 mol
e. 0.20 mol

I understand that C2H3O2- the weak base would have 0.05 moles found by multiplying (0.50 L) x (0.10 M) to get 0.05 moles. I know that acid and base react in a 1:1 ratio in buffers.

However, I am confused as to how the answer is B. I thought when you add a strong acid (HCl) to a weak base (C2H3O2-) you add half as much.

You can think as an acid-base stoichiometry

C2H3O2- + HCl -> HC2H3O2 + Cl-
The ratio is 1:1 so 0.05 mol base would react with 0.05 mol acid

About the weak base, even though the conjugate acid, acetic acid, is a strong acid with pKa around 4, it is still weaker than HCl (pKa <0). Therefore the reaction would favor the formation of acetic acid. I dont know where you got the idea of half ratio, but it is still 1:1 ratio.
Hope this would help... This is my first time answering here 🙂