Question
A chemistry graduate student is given 125.mL of a 1.20M propanoic acid HC2H5CO2 solution. Propanoic acid...
A chemistry graduate student is given 125.mL of a 1.20M propanoic acid HC2H5CO2 solution. Propanoic acid is a weak acid with =Ka×1.310−5 What mass of KC2H5CO2 should the student dissolves in the HC2H5CO2 solution to turn it into a buffer with pH =5.13
Be sure your answer has a unit symbol, and round it to 2 significant digits.
?
Answers
Ka = 1.3*10^-5
pKa = - log (Ka)
= - log(1.3*10^-5)
= 4.886
use formula for buffer
pH = pKa + log ([KC2H5CO2]/[HC2H5CO2])
5.13 = 4.8861 + log ([KC2H5CO2]/[HC2H5CO2])
log ([KC2H5CO2]/[HC2H5CO2]) = 0.2439
[KC2H5CO2]/1.2 = 1.7537
[KC2H5CO2] = 2.1044
volume , V = 1.25*10^2 mL
= 0.125 L
use:
number of mol,
n = Molarity * Volume
= 2.104*0.125
= 0.263 mol
Molar mass of KC2H5CO2,
MM = 1*MM(K) + 3*MM(C) + 5*MM(H) + 2*MM(O)
= 1*39.1 + 3*12.01 + 5*1.008 + 2*16.0
= 112.17 g/mol
use:
mass of KC2H5CO2,
m = number of mol * molar mass
= 0.263 mol * 1.122*10^2 g/mol
= 29.51 g
Answer: 30. g
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