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andrezito [222]
1 year ago
13

Calculate the concentration (M) of sodium ions in a solution made by diluting 10.0 mL of a 0.774 M solution of sodium sulfide to

a total volume of 100 mL.
Chemistry
1 answer:
Luda [366]1 year ago
6 0

The concentration of sodium ions in a solution made by diluting 10.0 mL of a 0.774 M solution of sodium sulfide to a volume of 100 mL would be 0.1548 M.

<h3>Dilution</h3>

According to the dilution principle, the number of moles of solutes in a solution before and after dilution must be the same. This is expressed as the following equation:

m_1v_1 = m_2v_2

Where m_1 is the molarity before dilution, v_1 is the volume before dilution, m_2 is the molarity after dilution, and v_2 is the volume after dilution.

In this case, m_1 = 0.774 M, v_1 = 10.0 mL, v_2 = 100 mL.

m_2 = m_1v_1/v_2

     = 0.774 x 10/100

      = 0.0774 M

Thus, the new concentration of the sodium sulfide solution would be 0.0774 M.

Mole of the final solution: 0.0774 x 0.1 = 0.00774 mol

Sodium sulfide formula = Na_2S ---> 2Na^+ + S^{2-

Equivalent mole of sodium ion = 0.00774 x 2

                                                    = 0.01548 mol

The concentration of sodium ions = mol/volume

                                                  = 0.01548/0.1

                                                  = 0.1548 M

In other words, the concentration of the sodium ions in the diluted solution would be 0.1548 M.

More on dilution can be found here: brainly.com/question/21323871

#SPJ1

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What is the standard electrode potential for a galvanic cell constructed in the appropriate way from these two half-cells?
____ [38]

E

θ

Cell

=

+

2.115

l

V

Cathode

Mg

2

+

/

Mg

Anode

Ni

2

+

/

Ni

Explanation:

Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]

Mg

2

+

(

a

q

)

+

2

l

e

−

→

Mg

(

s

)

−

E

θ

=

−

2.372

l

V

Ni

2

+

(

a

q

)

+

2

l

e

−

→

Ni

(

s

)

−

E

θ

=

−

0.257

l

V

The standard reduction potential

E

θ

resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (

298

l

K

,

1.00

l

kPa

) is defined as

0

l

V

for reference. [2]

A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.

Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher

E

θ

and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower

E

θ

will experience oxidation and act the anode.

E

θ

(

Ni

2

+

/

Ni

)

>

E

θ

(

Mg

2

+

/

Mg

)

Therefore in this galvanic cell, the

Ni

2

+

/

Ni

half-cell will experience reduction and act as the cathode and the

Mg

2

+

/

Mg

the anode.

The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:

E

θ

cell

=

E

θ

(

Cathode

)

−

E

θ

(

Anode

)

E

θ

cell

=

−

0.257

−

(

−

2.372

)

E

θ

cell

=

+

2.115

Indicating that connecting the two cells will generate a potential difference of

+

2.115

l

V

across the two cells.

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