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garik1379 [7]
3 years ago
6

chemist must prepare of hydrochloric acid solution with a pH of at . He will do this in three steps: Fill a volumetric flask abo

ut halfway with distilled water. Measure out a small volume of concentrated () stock hydrochloric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrochloric acid that the chemist must measure out in the second step. Round your answer to significant digits.
Chemistry
1 answer:
olya-2409 [2.1K]3 years ago
3 0

Answer:

1.7mL of the stock solution

Explanation:

<em>...Must prepare 550.0mL with a pH of 1.60 at 25°C</em>

<em>...Fill a 550.0mL volumetric flask</em>

<em>...Concentrated stock of 8.0M</em>

<em>... Round the answer to 2 significant digits</em>

<em />

First, with pH we can solve [H⁺] = [HCl] concentration that we need to prepare the solution:

pH = -log[H⁺]

1.60 = -log[H⁺]

10^-1.60 = [H⁺]

0.0251M = [H⁺]

That means we need to dilute the concentrated solution from 8M to 0.0251M, that is:

8M / 0.0251M = 318.5 times

As the final volume of the solution is 550.0mL, the volume of the stock 8M HCl solution must be:

550.0mL / 318.5 times =

<h3>1.7mL of the stock solution</h3>

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(b) The cell potential when the concentration of Ni^{2+} has fallen to 0.500 M is, 0.52 V

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From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : Zn\rightarrow Zn^{2+}+2e^-     E^0_{[Zn^{2+}/Zn]}=-0.76V

Reaction at cathode (reduction) : Ni^{2+}+2e^-\rightarrow Ni     E^0_{[Ni^{2+}/Ni]}=-0.23V

The balanced cell reaction will be,  

Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)

First we have to calculate the standard electrode potential of the cell.

E^o=E^o_{cathode}-E^o_{anode}

E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}

E^o=(-0.23V)-(-0.76V)=0.53V

(a) Now we have to calculate the cell potential.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}

where,

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

E_{cell} = emf of the cell = ?

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

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}

E_{cell}=0.49V

(b) Now we have to calculate the cell potential when the concentration of Ni^{2+} has fallen to 0.500 M.

New concentration of Ni^{2+} = 1.50 - x = 0.500

x = 1 M

New concentration of Zn^{2+} = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}

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

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}

E_{cell}=0.52V

(c) Now we have to calculate the concentrations of Ni^{2+} and Zn^{2+} when the cell potential falls to 0.45 V.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}

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

0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}

x=1.49M

The concentration of Ni^{2+} = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of Zn^{2+} = 0.100 + x = 0.100 + 1.49 = 1.59 M

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