Prelaboratory Assignment: pH and Buffers Calculate the expected pH of solutions
ID: 1046733 • Letter: P
Question
Prelaboratory Assignment: pH and Buffers Calculate the expected pH of solutions prepared as given below. Enter the results into the Calculated pH blocks on the second page of the report sheet. Buffer containing 4 mL 6 M acetic acid (pK=4.74) + 3.3 g sodium acetate trihydrate (MM = 136 g/mol) + water to make up 50 mL. Half (25 mL) of the buffer+ 6.0 mL 1 M HCl. The other half of the buffer + 6.0 mL 1 M NaOH. To the buffer with HCl add another 10 mL HCl for a total of 16.0 mL 1 M HCl. To the buffer with NaOH add another 10 ml NaOH for a total of 16.0 mL 1 M NaOH.Explanation / Answer
Sol:-
(a) pH of 50ml Buffer solution :-
number of moles of CH3COOH = 6M x 0.004L = 0.024 mol ( because, molarity = number of moles / vol. of solution in Litre)
and number of moles of C2H9NaO5 = 3.3g / 136g/mol = 0.024 mol ( because,number of moles = given mass in g / gram molar mass )
Now total vilume of the solution becomes = V = 0.050L + 0.004L = 0.054L
therefore, by using Henderson -Hasselbalch equation , we have
pH = pka + log [salt] / [ acid ] ..............(1)
pH = 4.74 + log [ C2H9NaO5] / [CH3COOH]
pH = 4.74 + log 0.024M / 0.024M ( because , [ C2H9NaO5] = 0.024 / 0.054 and [CH3COOH] = 0.024/0.054)
pH = 4.74 + log 1
pH = 4.74 + 0 ( beacause log1 = 0)
pH = 4.74
(b) 6.0mL of 1M HCl :-
number of moles of HCl = 1M x 0.006 L = 0.006 mol
therefore, reaction of HCl with buffer will be
HCl + C2H9NaO5 ---------> C2H9O5H (acid ) + NaCl
I 0.006mol 0.024mol 0.024mol -
C -0.006 -0.006 +0.006
F 0 0.0018mol 0.03mol
again from equation (1) we have
pH = 4.74 + log [ C2H9NaO5] / [C2H9O5H]
pH= 4.74 + log 0.018 / 0.03
pH = 4.74 + log 0.6
pH = 4.74 - 0.2218
pH = 4.518
(c) 6.0mL of 1M NaOH:-
number of moles of NaOH = 1M x 0.006 L = 0.006 mol
therefore reaction of NaOH with buffer will be
NaOH + CH3COOH ---------> CH3COONa + H2O
I 0.006mol 0.024mol 0.024mol -
C -0.006 -0.006 +0.006
F 0 0.0018mol 0.03mol
again from equation (1) we have
pH = 4.74 + log [ CH3COONa] / [CH3COOH]
pH= 4.74 + log 0.03 / 0.018
pH = 4.74 + log 1.66
pH = 4.74 + 0.22
pH = 4.96
(d) 16.0mL of 1M HCl :-
number of moles of HCl = 1M x 0.016 L = 0.016mol
therefore reaction of HCl with buffer will be
HCl + C2H9NaO5 ---------> C2H9O5H (acid ) + NaCl
I 0.016mol 0.024mol 0.024mol -
C -0.016 -0.016 +0.016
F 0 0.008mol 0.04mol
again from equation (1) we have
pH = 4.74 + log [ C2H9NaO5] / [C2H9O5H]
pH= 4.74 + log 0.008 / 0.04
pH = 4.74 + log 0.2
pH = 4.74 - 0.6989
pH = 4.04
e). and similarly for 16.0ml of 1M NaOH
pH= 4.74 + log 0.04 / 0.008
pH = 4.74 + log 5
pH = 4.74 + 0.6989
pH = 5.438