Question

How many grams of sodium acetate you should dissolve in 100 mL to have a concentration of hydronium ions equal to 10^-9M (Ka = 1.8E-5)

Answer 1

[H^+]    = 1*10^-9M

[OH^-]   = Kw/[H^+]

             = 1*10^-14/(10^-9)   = 10^-5 M

Kb   = Kw/Ka

        = 1*10^-14/(1.8*10^-5) = 5.6*10^-10

CH3COONa(aq) -----------> CH3COO^- (aq) + Na^+(aq)

x ------------------------------------ x

------ CH3COO^- (aq) + H2O(l) ------------> CH3COOH(aq)   + OH^- (aq)

I -------- x------------------------------------------------ 0 --------------------- 0

C------ -10^-5 ------------------------------------------- 10^-5 ---------------- 10^-5

E ----- x-10^-5 ----------------------------------------- 10^-5 ---------------- 10^-5

               Kb    = [CH3COOH][OH^-]/[CH3COO^-]

               5.6*10^-10   = 10^-5*10^-5/(x-10^-5)

               5.6*10^-10(x-10^-5)   = 1*10^-10

              x = 0.178

CH3COONa(aq) -----------> CH3COO^- (aq) + Na^+(aq)

0.178M ------------------------------- 0.178M

no of moles of CH3COONa   = molarity*volume in L

                                          = 0.178*0.1 = 0.0178moles

mass of CH3COONa = no of moles * gram molar mass

                                 = 0.0178*82   = 1.46g >>>>answer