A 1-liter aqueous solution is prepared at 25 degree C with 10^-4 moles of acetic

Question

A 1-liter aqueous solution is prepared at 25 degree C with 10^-4 moles of acetic acid (HAc) and 10^-3 moles of calcium carbonate (CaCO_3) and reaches equilibrium. List the eight unknown chemical species here (water is not unknown). What pH is required to reduce a high concentration of a dissolved copper, Cu^2+, to 45 mg/L, by precipitation to Cu(OH)_2? The solubility product for the reaction is 7.8 times 10^-20. Nitrogen dioxide (NO_2) concentrations are measured in an air-quality study and decrease from 5 ppmv to 2 ppm in 4 min with a particular light intensity, (a) What is the first-order rate constant for this reaction? (b) What is the half-life of NO_2 during this study? Chick's law (given below) is often used to mathematically describe the inactivation (death) of microorganisms during chlorine disinfection as part of the overall drinking water treatment process. Integrate Chick's law, clearly showing all steps, to derive a mathematical expression to determine the number of microorganisms remaining (after a certain disinfection time), N_t, when given the initial number of microorganisms present, N_0. dN/dt = - kN where N is the number of microorganisms present k is the first order rate constant

Explanation / Answer

6) Calcium carbonate and Acetic acid reaction.

CH3COOH + CaCO3 = H2CO3 + Ca(CH3COO)2.

Again; carbonic acid breaks down to form carbon dioxide and water,
H2CO3 + H2O = HCO3- + H3O+
HCO3- + H2O = CO32- + H3O+
H2CO3 = H2O + CO2.

The overall reaction is the sum of the two reactions given,

2CH3COOH + CaCO3 = H2O + CO2 + Ca(CH3COO)2

Here, we have 10^-4 moles of HAC and 10^-3 moles of CaCO3. The HAC concentration is very low compared to CaCO3.

Here are the possible species present in this reaction:

CO2, Ca(CH3COO)2, H2CO3,HCO3-, CO32-,H3O+, CaCO3, CH3COOH

7) Write the dissociation equation:

    Cu(OH)2 = Cu2+ + 2OH¯

    Write the Ksp expression:

    Ksp = [Cu2+] [OH¯]2

   Plug into the Ksp expression:

    7.8 x 10¯20 = (0.045 g/L) (s)2

   Solve for [OH¯]:

    [OH¯] = s = 13.15 x 10¯10 M

   Calculate the pH given the [OH¯]:

    pOH = -log 13.15 x 10¯10 = 8.88
    pH = 14 - pOH = 14 - 8.88 = 5.12
  
8) a)for a first-order reaction; the rate constant is given by;

           k = 2.303/t log [A]o/[A]t

   [A]o = initial conc. = 5 ppmv

   [A]t = conc. at time t = 2 ppmv

   and t = 4 min

    Plug into the rate constant equation..

   k = 2.303/4 log 5/2
       = 2.303/4 log 2.5 = 0.229 min^-1

   b)half life of the first order reaction given by;
           t1/2= 0.693/k
               = 0.693/0.229 = 3.026 minutes

  
  

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