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bixtya [17]
3 years ago
11

Which of the following is not a property of a buffer solution? a) It is a mixture of a weak acid and its conjugate base. b) Resi

sts pH changes because it reacts with added acid or base. c) The maximum buffer capacity is at pH = pKa. d) pH is dependent on the solution ionic strength and temperature. e) pH is independent of concentration.
Chemistry
1 answer:
m_a_m_a [10]3 years ago
8 0

Answer:

e) pH is independent of concentration.

Explanation:

a) It is a mixture of a weak acid and its conjugate base. <em>TRUE. </em>A buffer is defined as a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid.

b) Resists pH changes because it reacts with added acid or base. <em>TRUE. </em>Thermodynamically, the reaction of added acid or base is faster with the buffer mixture than with H⁺ or OH⁻ ions of the solutions.

c) The maximum buffer capacity is at pH = pKa. <em>TRUE. </em>The buffer capacity is pka±1. For this, buffer capacity is maximum in pka.

d) pH is dependent on the solution ionic strength and temperature. <em>TRUE.</em> Ionic strength and temperature are factors that influence concentrations of ions in solutions as the H⁺ ion that is the responsible

e) pH is independent of concentration. <em>FALSE. </em>pH in a buffer depends completely of concentrations of the acid and its conjugate base or vice versa.

I hope it helps!

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Answer : The cell potential for this reaction is 0.50 V

Explanation :

The given cell reactions is:

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The half-cell reactions are:

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First we have to calculate the cell potential for this reaction.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 25^oC=273+25=298K

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = +0.63 V

E_{cell} = cell potential for the reaction = ?

[Zn^{2+}] = 3.5 M

[Pb^{2+}] = 2.0\times 10^{-4}M

Now put all the given values in the above equation, we get:

E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}

E_{cell}=0.50V

Therefore, the cell potential for this reaction is 0.50 V

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Explanation:

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Answer:

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