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ale4655 [162]
2 years ago
9

Which statement best describes the pH of pure water? It is neutral because the concentration of hydronium ions equals that of hy

droxide ions. It is neutral because the pure liquid contains neither hydronium ions nor hydroxide ions. It is acidic because it has a hydronium ion concentration of mc027-1. Jpg. It is basic because it has a hydroxide ion concentration of mc027-2. Jpg.
Chemistry
2 answers:
stira [4]2 years ago
4 0

The pH of pure water has been best described as neutral pH with equal hydronium and hydroxide ions. Thus, option A is correct.

pH has been described as the measurement of hydrogen ions in a solution. The pH has been measured on a scale of 1-14. pH 7 has been the neutral pH.

The higher hydronium ion concentration tends to move the pH from 7 towards 1. The higher hydroxide ion concentration tends to move the pH above 7.

The neutral pH has been neither acidic nor basic with the equal constituents of hydronium and hydroxide ion in the solution.

Thus, the pH of pure water has been 7. It has neutral pH with equal hydronium and hydroxide ions. Thus, option A is correct.

For more information about the pH of the solution, refer to the link:

brainly.com/question/4975103

GalinKa [24]2 years ago
3 0

Answer:

Water is neutral because there is an equal concentration of hydronium ions and hydroxide ions.

Explanation:

Water autoionizes into equal parts hydronium and hydroxide ions, resulting in a neutral pH. This is because during the autoionization reaction, two water molecules yield one hydronium and one hydroxide.

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The equilibrium constant is  K =0.02867

The equilibrium pressure for oxygen gas is  P_{O_2} =  0.09367\  atm

Explanation:

  From the question we are told that

       The equation of the chemical reaction is

                    M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}

   The Gibbs free energy forM_3 O_4_{(s)} is  \Delta G^o_{1}  = -8.80 \ kJ/mol

  The Gibbs free energy forM{(s)} is  \Delta G^o_{2}  = 0 \ kJ/mol

    The Gibbs free energy forO_2{(s)} is  \Delta G^o_{3}  = 0 \ kJ/mol

The Gibbs free energy of the reaction is mathematically represented as

         \Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r

         \Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)

Substituting values

From the balanced equation

         \Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]

        \Delta G^o_{re} = 8.80 kJ/mol =8800J/mol

The Gibbs free energy of the reaction can also be represented mathematically as

           \Delta G^o_{re} = -RTln K

Where R is the gas constant with a value of  R = 8.314 J/mol \cdot K

             T is the temperature with a given value  of  T = 298 K

             K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction  is mathematically represented as

                          K_p =[ P_{O_2}]^{\frac{3}{2} }

Where   [ P_{O_2}] is the equilibrium pressure of oxygen

         The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the  second equation of the Gibbs free energy of the reaction

 

           K = e^{- \frac{\Delta G^o_{re}}{RT} }

Substituting values

           K= e^{\frac{8800}{8.314 * 298} }

            K =0.02867

Now substituting this into the equation above to obtain the equilibrium of oxygen

           0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}

multiplying through by 1 ^{\frac{2}{3} }

        P_{O_2} =  [0.02867]^{\frac{2}{3} }

        P_{O_2} =  0.09367\  atm

       

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