How many moles of NaOH would need to be added to buffer to complete neutralization? Please help!! :(?

Question

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?

I got 0.024 moles

b. How many moles of NaOH would need to be added to the 19 mL of buffer to destroy the system? (complete neutralization)

This is where I am getting confused. Neutralize meaning that the pH would be brought to 14 for NaOH and 0 when adding HCL from part C? If yes, how do the calculations work? Explanation would be greatly appreciated.

c. How many mL of 2.0 M HCL would be needed to destroy 19 mL of the buffer system?
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Answer 1

b. 0.024 moles (the same as the acid).



c. Work out the moles of solid sodium ethanoate (3.3/148). Then multiply by 19/50. That is the number of moles of HCl required. Volume = 1,000 x moles/2.0.

Answer 2

b. HC2H3O2+H20 <-> H3O + C2H3O2

.024mol

-.024mol

Na+ OH-> NaOH



c. 3.3g (1mol/136g)= .024

.024(.019L/.050L)= .092mols

.092 x 1000ml= 92M

92M/2M= 46ml

Answer 3

the wonderful type of the equation could be (X + A)^2 = B. the middle term is 26X and if (X + A)(X + A) is greater effective, the end result's X^2 + 2AX + A^2. subsequently, 2AX = 26X and A = 13. X^2 + 26X + 169 = 33 + 169 (X + 13)^2 = 182