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Kryger [21]
3 years ago
6

Given pH = 3.50 Find: [H3O+] and [OH-] Is this acidic, basic or neutral?

Chemistry
1 answer:
lys-0071 [83]3 years ago
8 0

Answer:

Explanation:

Given parameters:

           pH = 3.50

Unknown:

    concentration of [H₃0⁺] = ?

    concentration of [OH⁻] = ?

Solution:

In order to find the unknown, we use some simple expressions which best explains the pH scale and the equilibrium systems of aqueous solutions.

         pH = -log₁₀[H₃O⁺]

         [H₃O⁺] = inverse log₁₀ (-pH) = 10^{-pH} = 10^{-3.5}

          [H₃O⁺] = 3.2 x 10⁻⁴moldm⁻³

       

For the  [OH⁻]:

       we use : pOH = -log₁₀ [OH⁻]

     Recall: pOH + pH = 14

                  pOH = 14 - pH = 14 - 3.5 = 10.5

  Now we plug the value of pOH into pOH = -log₁₀ [OH⁻]

                                   [OH⁻] = 10^{-pOH}

                        [OH⁻] = 10^{-10.5} = 3.2 x 10⁻¹¹moldm⁻³

The solution is acidic as the concentration of H₃0⁺ is more than that of the OH⁻ ions.

                   

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An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective
katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

PCl_3,p_1'=6.798 Torr

Cl_2,p_2'=26.398 Torr

PCl_5,p_3'=223.402 Torr

Explanation:

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At equilibrium before adding chlorine gas:

Partial pressure of the PCl_3=p_1=13.2 Torr

Partial pressure of the Cl_2=p_2=13.2 Torr

Partial pressure of the PCl_5=p_3=217.0 Torr

The expression of an equilibrium constant is given by :

K_p=\frac{p_1}{p_1\times p_2}

=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245

At equilibrium after adding chlorine gas:

Partial pressure of the PCl_3=p_1'=13.2 Torr

Partial pressure of the Cl_2=p_2'=?

Partial pressure of the PCl_5=p_3'=217.0 Torr

Total pressure of the system = P = 263.0 Torr

P=p_1'+p_2'+p_3'

263.0Torr=13.2 Torr+p_2'+217.0 Torr

p_2'=32.8 Torr

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

K_p=\frac{p_3'}{p_1'\times p_2'}

1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr

p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr

p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr

6 0
3 years ago
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WITCHER [35]

Answer:

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Explanation:

7 0
2 years ago
Why are atoms of carbon-14(C-14) unstable?
vfiekz [6]
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A solution is made by mixing 34.5 g of sugar with 75.0 g of water. What is the mass percent of sugar in this solution?
lubasha [3.4K]

the mass percent of sugar in this solution is 46%.

Answer:

Solution given:

mass of solute=34.5g

mass of solvent=75g

mass percent=\frac{mass\:of\:solute}{mass\:of \:solvent}*100\%

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student in chemistry 150-02 weighed out 55.5g of octane c8h18 and allowed it to react with oxygen, o2 the product formed were ca
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Answer:

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(2) 10(4H+ + O2 → 2H2O)

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(1) + (2):

balanced reaction:

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