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ELEN [110]
3 years ago
7

Calculate the poH if 2.95 g of KOH(s) dissolved to make 100mL of solution

Chemistry
1 answer:
Elodia [21]3 years ago
4 0

Answer:

\rm pOH \approx 1.279 if 2.95\; \rm g of \rm KOH is dissolved completely in water to make a 100\; \rm mL solution.

Explanation:

Look up the relative atomic mass data on a modern periodic table:

  • \rm K: 39.098.
  • \rm O: 15.999.
  • \rm H: 1.008.

Calculate the formula mass of \rm KOH:

\begin{aligned}&M(\mathrm{KOH})\\ &= 39.098 + 15.999 + 1.008 \\ &= 56.105\; \rm g \cdot mol^{-1}\end{aligned}.

Find the number of moles of formula units in that 2.95\; \rm g of \rm KOH:

\begin{aligned}& n(\mathrm{KOH}) \\ & = \frac{m(\mathrm{KOH})}{M(\mathrm{KOH})} = \frac{2.95\; \rm g}{56.105\; \rm g \cdot mol^{-1}} \approx 0.0525800\; \rm mol \end{aligned}.

\rm KOH is a strong base. When dissolved in water, it ionizes completely to produce \rm K^{+} ions and hydroxide ions (\rm OH^{-}.)

Note that there are one moles of \rm OH^{-} ions in each mole of \rm KOH formula units. When that 2.95\; \rm g of \rm KOH (approximately 0.0525800\; \rm mol of \rm KOH formula units) dissolves completely in water to make a 100\; \rm mL solution, about 0.0525800\; \rm mol of \rm OH^{-} ions will be produced.

Convert the unit of volume to liters: V = 100\; \rm mL = 0.1\; \rm L.

Calculate the concentration of \rm OH^{-} ions in that solution:

\begin{aligned}&[\mathrm{OH^{-}}] \\ &= \frac{n}{V} \approx \frac{0.0525800\; \rm mol}{0.1 \; \rm L} \approx 0.525800\; \rm mol \cdot L^{-1} \end{aligned}.

Calculate the \rm pOH of that solution:

\begin{aligned}& \mathrm{pOH} \\ &= -\log_{10}[\mathrm{OH^{-}}] \\ &\approx -\log_{10}(0.525800) \approx 1.279\end{aligned}.

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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

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=

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l

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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]

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(

s

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s

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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

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1. B = 1.13M

2. A. 8.46%

3. D = 0.0199

Explanation:

1. Molarity of a a solution = number of moles of solute/ volume of solution in L

Number of moles of solute = mass of solute/molar mass of solute

Molar mass of NaC₂H₃O₂ = 82 g/mol, mass of NaC₂H₃O₂ = 245 g

Number of moles of NaC₂H₃O₂ = 245 g / 82 g/mol = 2.988 moles

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Density of water = 1 Kg/L = 1000 g/L; volume of waterb= 2.65 L

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3. Mole fraction of NaC₂H₃O₂ = moles of NaC₂H₃O₂/moles of solution

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Moles of NaC₂H₃O₂ =2.988 moles

Moles fraction of NaC₂H₃O₂ =2.988/150.21 = 0.0199

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