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olga_2 [115]
3 years ago
10

Find the theoretical oxygen demand for the

Chemistry
1 answer:
never [62]3 years ago
7 0

Answer:

a) 213.3 mg/L

b) 62.61 mg/L

c) 0.0225 mg/L

Explanation:

Theoretical oxygen demand (ThOD)is essentially the amount of oxygen required for the complete degradation of a given compound into the final oxidized products

a) Given:

Concentration of acetic acid,[CH3COOH] = 200 mg/L

CH3COOH + 2O2 \rightarrow 2CO2 + 2H2O

ThOD = \frac{mass\ O2}{mass\ CH3COOH} * conc. CH3COOH

Based on the reaction stoichiometry:

mass of CH3COOH = 60 g

mass of O2= 2(32) = 64 g

ThOD = \frac{64 g}{60 g} * 200mg/L = 213.3 mg/L

b) Given:

Concentration of ethanol, [C2H5OH] = 30 mg/L

C2H5OH + 3O2 \rightarrow 2CO2 + 3H2O

ThOD = \frac{mass\ O2}{mass\ C2H5OH} * conc. C2H5OH

Based on the reaction stoichiometry:

mass of C2H5OH = 46 g

mass of O2= 3(32) = 96 g

ThOD = \frac{96 g}{46 g} * 30mg/L = 62.61 mg/L

c) Given:

Concentration of sucrose, [C12H22O11] = 50 mg/L

C12H22O11 + 12O2 \rightarrow 12CO2 + 11H2O

ThOD = \frac{mass\ O2}{mass\ C22H22O11} * conc. C12H22O11

Based on the reaction stoichiometry:

mass of C12H22O11 = 342 g

mass of O2= 12(32) = 384 g

ThOD = \frac{384 g}{342 g} * 50mg/L = 0.0225 mg/L

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Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1
NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x                       x               x

The dissociation constant from the above reactions is given by:

Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}

6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}

6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75    

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

\% = \frac{x}{[HCN]} \times 100

\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100

\% = 3.5 \cdot 10^{-3} \%          

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!      

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