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MAVERICK [17]
2 years ago
14

The student adds 0.0010 mol of NaOH(s) to solution Y, and adds 0.0010 mol of NaOH(s) to solution Z. Assume that the volume of ea

ch solution does not change when the NaOH(s) is added. The pH of solution Y changes much more than the pH of solution Z changes. Explain this observation.
Chemistry
1 answer:
julia-pushkina [17]2 years ago
6 0

Answer:

The answer is in the explanation.

Explanation:

A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa. Buffers are able to avoid the pH change of a solution when strong acid or bases are added (As NaOH).

Based on the experiment, it is possible that the solution Z was a buffer and Y another kind of solution. For this reson, pH of the solution Y changes much more than the pH of solution Z changes despite the amount of NaOH added is the same in both solutions.

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Using the derived expression from Arrhenius Equation

In \ (\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})

Given that:

time t_1 = 8.3 days = (8.3 × 24 ) hours = 199.2 hours

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Temperature T_2 = 30° C = (30+ 273) = 303 K

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Since (\frac{k_2}{k_1}=\frac{t_2}{t_1})

Then we can rewrite the above expression as:

In \ (\frac{t_2}{t_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})

In \ (\frac{199.2}{10.6}) = \frac{E_a}{8.314}(\frac{303-273}{273*303})

2.934 = \frac{E_a}{8.314}(\frac{30}{82719})

2.934 = \frac{30E_a}{687725.766}

30E_a = 2.934 *687725.766

E_a = \frac{2.934 *687725.766}{30}

E_a =67255.58 \ J/mol

E_a =67.2 \ kJ/mol

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