a. The pKa of acetic acid is 4.7. What is the buffering range of this acid and i
ID: 1008996 • Letter: A
Question
a. The pKa of acetic acid is 4.7. What is the buffering range of this acid and its conjugate base? Explain your answer. b. Calculate the volume of 6 M acetic acid needed to prepare 100 mL of a 0.10 M acetic acid (CH3COOH) solution. c. Calculate the mass of sodium acetate (CH3COONa) required to prepare 50 mL of a 0.10 M sodium acetate solution. d. Calculate the volume of 1 M sodium hydroxide (NaOH) needed to prepare 10 mL of a 0.10 M sodium hydroxide solution. e. Calculate the volume of 1 M hydrochloric acid (HCl) needed to prepare 10 mL of a 0.10 M hydrochloric acid solution. f. Determine the volumes of 0.10 M CH3COOH and 0.10 M CH3COONa required to prepare 10 mL of the following pH buffers. (Note: the pKa of CH3COOH =4.7) a. pH 3.7 b. pH 4.7 c. pH 5.7
Explanation / Answer
Buffer solution resists change of pH to the capacity of +/- 1.
pH=pka+/- 1
B) M1*V1=M2*V2
M1=6M
V1=?
M2=0.10M
V2=100 ml
V1=M2 V2/M1=0.10*100/6=1.7 ml
C) moles of sodium acetate (CH3COONa) required=molarity*volume=0.1 mol/L*0.050 L=0.005 moles
Mass of sodium acetate (CH3COONa) reqd=0.005 moles*molar mass=0.005 mol*82.02g/mol=0.410g
D)M1 *V1=M2*V2
M1=0.1M
V1=10ml
M2=1M
V2=M1V1/M2=0.1 M*10ml/1M=1 ml
E)M1 *V1=M2*V2
M1=0.1M
V1=10ml
M2=1M
V2=M1V1/M2=0.1 M*10ml/1M=1 ml
F)a) pH=pka+log[base]/[acid] [henderson -hasselbach eqn)
3.7=4.7 +log[base]/[acid]
log[base]/[acid]=-10
[base]/[acid]=10^-10
0.10M*Vb/0.10M*Va=10^-10
Vb/Va=10^-10
Va+Vb=10 ml
Va=1/(1+10^-10) *10 ml=10 ml (almost)
Vb=10^-10/(1+10^-10)*10ml=10^-10 ml
B)4.7=4.7+llog[base]/[acid]
0=log[base]/[acid]
10^0=1=[base]/[acid]
[base]/[acid]=1
[base]=[acid]
0.1M*Va=0.1 M*Vb
Va=Vb=5ml
C) 5.7=4.7+log[base]/[acid]
10=og[base]/[acid]
[base]/[acid]=10^10
Vb/Va=10^10
VB=10^10/(1+10^10)*10 ml=10ml (approx)
Va=1/(1+10^10*10ml)*10ml=10^-9 ml
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