The [H3O+] in a solution is increased to twice the original concentration. Which change could occur in the pH? 2.0 to 4.0 1.7 to 1.4 5.0 to 2.5 8.5 to 6.5 6.8 to 9.8

Respuesta :

2.0 to 4.0 the way to figure it out is right in the problem


Answer : The correct option is, 1.7 to 1.4

Explanation :

As we know that,

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

The formula of pH is:

[tex]pH=-\log [H_3O^+][/tex]

Let the initial concentration of hydronium ion be, 'X'

The expression of pH at initial concentration will be,

[tex]pH=-\log [H_3O^+]=-\log (X)[/tex]

According to the question, the concentration of hydronium ions is increased to twice the original concentration. That means,

[tex][H_3O^+]=X[/tex]

[tex][H_3O^+]_{new}=2\times [H_3O^+][/tex]

[tex][H_3O^+]_{new}=2X[/tex]

The expression of new pH will be,

[tex]pH_{new}=-\log [H_3O^+]=-\log (2X)[/tex]

[tex]pH_{new}=-\log (2X)=-\log 2-\log (X)[/tex]

[tex]pH_{new}=-\log 2-pH[/tex]

And the value of [tex]\log 2=0.3[/tex]. Now put this value in above expression, we get

[tex]pH_{new}=-0.3-pH[/tex]

This means that the new pH will decrease by a value of 0.3 and from the given options, the difference in pH corresponds to 1.7 to 1.4.

Hence, the correct option is, 1.7 to 1.4

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