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53) A buffer solution contains carbonic acid (H CO) and sodium bi carbonate (NaHCO), each at a concentration of 0.100 M. The
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53. Given the Ka of carbonic acid (H2CO3) is Ka = 4.5*10-7,

pKa = -log (Ka)

= -log (4.5*10-7)

= 6.347

Use the Henderson-Hasslebach equation to determine the pH.

pH = pKa + log [NaHCO3]/[H2CO3]

====> pH = 6.347 + log (0.1 M)/(0.1 M)

====> pH = 6.347 + log (1.0)

====> pH = 6.347 + 0.0 = 6.347

====> pH ≈ 6.3 (ans).

54. The pH of water is 6.77. Therefore,

pH = -log [H3O+] = 6.77

====> log [H3O+] = -6.77

====> [H3O+] = antilog (-6.77)

====> [H3O+] = 1.698*10-7

The hydronium ion concentration in water is 1.698*10-7 M.

Consider the auto-ionization of water as below.

2 H2O (l) <======> H3O+ (aq) + OH- (aq)

Due to the 1:1 nature of ionization,

[H3O+] = [OH-] = 1.698*10-7 M.

The ion product of water is given as

Kw = [H3O+][OH-]

= (1.698*10-7 M)*(1.698*10-7 M)

= 2.883*10-14 M2 ≈ 2.88*10-14 M2

The ion product is expressed as a pure number; hence, the ion product of water at 40ºC is 2.88*10-14 (ans).

55. Use the Henderson-Hasslebach equation.

pH = pKa + log [CH3CO2-]/[CH3CO2H]

Plug in values and get

2.74 = 4.74 + log [CH3CO2-]/[CH3CO2H]

=====> -2.00 = log [CH3CO2-]/[CH3CO2H]

=====> [CH3CO2-]/[CH3CO2H] = antilog (-2.00)

=====> [CH3CO2-]/[CH3CO2H] = 0.01 (ans).

56. The equilibrium constant for the reaction is given as

K = [NH3]2/[N2][H2]3

= (7.62 M)2/(4.53 M)(2.49 M)3

= 0.830 M-2

The equilibrium constant is expressed as a dimensionless quantity; hence, the equilibrium constant is 0.830 (ans).

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53) A buffer solution contains carbonic acid (H CO) and sodium bi carbonate (NaHCO), each at a concentration of 0.100 M. The relevant equilibrium is shown below. What is H:COs(ag)+ HOo (a)+HCOs (...
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