(6 pts) 19. A buffer is composed of HC,HsO2 and NaC,Hs0 2. Write the balanced ch
ID: 1044534 • Letter: #
Question
(6 pts) 19. A buffer is composed of HC,HsO2 and NaC,Hs0 2. Write the balanced chemical equation for the addition of HBr to this buffer. Also, write the balanced chemical equation for the addition of LiOH to the buffer. (10 pts) 20. What is the pH of a solution prepared by dissolving 4.1 g of sodium acetate (NaC2H302) in one liter of water. For acetic acid (HC2H302), Ka 1.8 x 10-5 (12 pts) 21. Describe how you would prepare one liter of a nitrite (NO2") buffer with a pH of 8.00. Ka 4.5 x 104 (for HNO,)Explanation / Answer
19.
HC7H5O2 ---------- H+ + C7H5O2-
NaC7H5O2 ----------- Na+ + C7H5O2-
The buffer solution contains C7H5O2- , H+ and Na+ ions.
By adding HBr,
HBr---------- H+ + Br-
The formed H+ ions from HBr combines with C7H5O2- in solution to form undissociated weak acid HC7H5O2 .
so there is no change in PH of the solution
C7H5O2- + H+ -------------- HC7H5O2
By adding LiOH,
LiOH---------------- Li+ + OH-
The formed OH- ions fromLiOH combines with H+ ions in solution to form undissociated water molecule.
so there is no change in PH of the solution.
H+ + OH- ------------- H2O
20)
mass of sodium acetate = 4.1 grams
molar mass of sodium acetate = 82.0 gram/mole
number of moles of sodium acetate = 4.1/82.0 = 0.05 moles
volume = 1L
Molarity = number of moles/volume= 0.05/1.0= 0.05M
Molarity = 0.05M
NaCH3COO = 0.05M
NaCH3COO----------------- Na+ + CH3COO-
CH3COO- H2O ----------- CH3COOH + OH-
0.05 0 0
-x +x +x
0.05-x +x +x
Ka= 1.8x10^-5
KaxKb= Kw where Kw = ionic product of water= 1.0x10^-14
Kb= Kw/Ka= 1.0x10^-14/1.8x10^-5 =5.56x10^-10
Kb = [CH3COOH][OH-]/[CH3COO-]
5.56x10^-10 = x*x/(0.05-x)
for solving the eqaution
x= 5.27x10^-6
[OH-]= 5.27x10^-6M
-log[OH-]= -log( 5.27x10^-6)
POH=5.278
PH+POH=14
PH= 14-POH
PH= 14-5.278
PH=8.722
PH= 8.72
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