Question

	Buffer A	Buffer B
Mass of NaC2H3O2 used to prepare buffer (g)	.149	1.49
Volume of buffer prepared (mL)	100.0	100.0
Molar concentration of HC2H3O2 in buffer (M)	0.1	1.0
Initial pH of buffer	4.0	4.0
Volume of 0.5 M NaOH to raise pH by 2 units (mL)	2	5
Volume of 0.5 M HCl to lower pH by 2 units (mL)	3	7
Volume of 0.5 M NaOH at equivalence point (mL)	4.5	10

1. Buffer capacity has a rather loose definition, yet it is an important property of buffers. A commonly seen definition of buffer capacity is: “The amount of H⁺ or OH^– that can be neutralized before the pH changes to a significant degree.” Use your data to determine the buffer capacity of Buffer A and Buffer B when a strong acid is added to it—that is, how many moles of acid can be added before the buffer is depleted? How many moles of base can be added before the buffer is depleted? Show this work in your “Data and Calculations” section.

2. In your “Results” section you should report the initial pH of each buffer you made as well as the buffering capacity of each solution.

Answer 1

INITIAL PKa of HC2H3O2=4.74   

Mass of salt (NaC2H3O2)=0.149gm ;; Molar mass of NaC2H3O2=82;; [NaC2H3O2]=(0.149$\div$82)$\div$(100$\div$1000)=0.018M;;;; [HC2H3O2]=0.1M;; moles of weak acid  HC2H3O2 in 100 ml buffer solution=(100$\div$1000)$\times$0.1=0.01mol.; Buffer is mixture of (HC2H3O2+ NaC2H3O2)

PH=PKa+LOG[NaC2H3O2]$\div$[HC2H3O2];;; PH=4+LOG(0.018$\div$0.1)=4;

AFTER ADDING 3ml of 0.5M HCl is added to buffer (HC2H3O2+ NaC2H3O2) then acid react with salt(NaC2H3O2) part and form acid(HC2H3O2)

moles of HCl =0.0015mol;;; moles of salt(NaC2H3O2)=0.0018mol ;;;; salt left after reaction=0.0003mol;; moles of  weak acid formed after reaction =0.0015mol.;;;; total moles of weak acid=0.01+0.0015=0.0115mol.;;

PH=PKa+log[NaC2H3O2]$\div$[HC2H3O2]=4.74+log0.0003$\div$0.0115=4.74-log115$\div$3=2(approx)

so change in PH=2

AFTER ADDING 2ml of 0.5M NaOH is added to buffer solution then strong base react with weak acid and form salt

moles of NaOH=0.001mol;; moles of weak acid(HC2H3O2)=0.01;; moles of weak acid left afer reaction=0.009;;; moles of salt(NaC2H3O2) formed after reaction=0.001mol;; total moles of salt=0.0010+0.0018=0.0028

PH=PKa+log[NaC2H3O2]$\div$[HC2H3O2]=4.74+log0.0028$\div$0.0090=2(approx)

change in PH=2 UNIT(APPROX)

WHEN 0.01mol strong acid and 0.0018mol of strong base added to this buffer solution then buffer is depleted and destroy.