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3 years ago

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Which statement is true for a solution when its concentration of hydroxide ions becomes equal to the concentration of hydronium

ions?
It becomes more acidic.

It becomes more alkaline.

Its pH becomes equal to 0.

Its pH becomes equal to 7.

2 answers:

bazaltina [42]3 years ago

7 0

Answer:

Option 4th, TRUE . Its pH becomes equal to 7.

Explanation:

[H₃O⁺] = [OH⁻]

pH = pOH → 7

[H₃O⁺] > [OH⁻] → pH < 7 ⇒ acidic solution

[H₃O⁺] < [OH⁻] → pH > 7 ⇒ basic solution

pH = 0 does not exist.

Greeley [361]3 years ago

3 0

Answer : The correct statement is, Its pH becomes equal to 7.

Explanation :

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

Mathematically,

$pH=-\log [H^+]$

When the pH less than 7 then the solution is acidic and the concentration of hydrogen ion is greater than hydroxide ion.
When the pH more than 7 then the solution is basic and the concentration of hydrogen ion is less than hydroxide ion.
When the pH is equal to 7 then the solution is neutral and the concentration of hydrogen ion is equal to the hydroxide ion.

As per question, the concentration of hydroxide ions becomes equal to the concentration of hydronium ions then its pH becomes equal to 7.

Hence, the correct statement is, Its pH becomes equal to 7.

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MariettaO [177]

Answer : The value of $K_c$ of the reaction is 10.5 and the reaction is product favored.

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First we have to calculate the concentration of $CH_3OH,CO\text{ and }H_2$ at equilibrium.

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Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

The expression of equilibrium constant $K_c$ for the reaction will be:

$K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$

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

$K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}$

$K_c=10.5$

Therefore, the value of $K_c$ of the reaction is, 10.5

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.

When $K_{c}$ ; the reaction is reactant favored.

When $K_{c}=1$ ; the reaction is in equilibrium.

As the value of $K_{c}>1$ . So, the reaction is product favored.

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