How do you calculate moles and neutralization in a buffer solution?

[qpham26]

Homework Statement

this is a part of a buffer solutions problem

3.3g of solid NaC2H3O2*3H2O added to 46mL of deionized water and 4.0 mL of 6.0M HC2H3O2.

a. How many moles of HC2H3O2 are contained in a 19 mL of the above buffer?
b. How many moles of NaOH would need to be added to the 19 mL of buffer to destroy the system? (complete neutralization)
c. How many mL of 2.0 M HCL would be needed to destroy 19 mL of the buffer system?

Homework Equations

The Attempt at a Solution

a. The concentration of HC2H3O2 will be 4x6/50 = 0.48M
moles of HC2H3O2 are contained in a 19ml will be 0.48M x 0.019mL = 9.12 x 10^-3 mols
b. same amount of moles as the acid?

c. clueless.


thanks for your time

 

[Borek]

Think what would be the reaction in c. Question is very similar to b, it is just a matter of realizing what is going on and following the stoichiometry.

 

[qpham26]

so to completely destroy the buffer system, the amount of HCl should be equal to the number of moles of NaC2H3O2*3H2O?
3.3/(MM of NaC2H3O2*3H2O) = 0.024 moles?
the volume will then be
0.024 mole / 2.0 M = 0.012L = 12mL

is this correct and also were the answer for a and b correct as well?

thanks for your time.

 

[Borek]

I have not checked the numbers, but the logic looks OK.