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Nesterboy [21]
2 years ago
5

In which of the following are the pH values arranged from the most basic to the most acidic?

Chemistry
1 answer:
MAVERICK [17]2 years ago
8 0

Answer:

The correct answer is : 14, 10, 7, 4, 3, 1

Explanation:

The pH of the solution is a negative logarithm of the hydronium ion's concentrations present in a aqueous solution.The pH scale ranges from 0 to 14.

Mathematically written as:

pH=-\log[H_3O^+]

  • Higher the value of pH more alkaline will be the solution.
  • Lower the value of pH more acidic will be the solution.
  • The solution with pH value 7 is termed as neutral solution.

So, from the given options of pH arranged from most basic to most acidic will be the decreasing order from 14 to 1 :

14, 10, 7, 4, 3, 1

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At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o
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Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

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The partial pressure of Br_2 = 2.0\times 10^{-4}atm

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The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

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3 years ago
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