Question

O What is the amount (in grams of sodium Acetate (MW = 82 g/l) To be added to 500.0 ml f 0-200 M acetic Acid (Ka= 1.8oxlo in

would you show me the step by step to solve this problem, thank you so much

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Answer #1

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

use formula for buffer

pH = pKa + log ([CH3CO2Na]/[CH3CO2H])

5.0 = 4.7447 + log ([CH3CO2Na]/[CH3CO2H])

log ([CH3CO2Na]/[CH3CO2H]) = 0.2553

[CH3CO2Na]/0.2 = 1.8

[CH3CO2Na] = 0.36

volume , V = 5*10^2 mL

= 0.5 L

use:

number of mol,

n = Molarity * Volume

= 0.36*0.5

= 0.18 mol

Molar mass of CH3CO2Na = 82 g/mol

use:

mass of CH3CO2Na,

m = number of mol * molar mass

= 0.18 mol * 82 g/mol

= 14.8 g

Answer: c

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