A solution has a hydronium-ion concentration of 1.0 x 10-4 mol per liter. What is the pH of this solution? Is this solution acidic or basic? What is the hydroxide-ion concentration of this solution?

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jerichorayel's profile pic

jerichorayel | College Teacher | (Level 2) Senior Educator

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pH is a measure of the concentration of the hydronium ions present in the solution. To solve for the pH of a particular solution, we can use:

`pH = -log[Acid] = -log[H^+]`

`pH = -log (1.0x10^(-4)) = 4`

If the pH of the solution is 7, the solution is neutral. If it is less than 7, it is an acidic. If it is more than 7, the solution is basic. Since the solution has the pH of 4, it is an acidic solution.

To get the hydroxide concentration of the solution, we will use the relationship:

`14 = pH + pOH `

where pOH is the measure of the hydroxide concentration. 

`pOH = 14 - 4 = 10 `

`[OH^(-)] = 10^(-pOH) = 10^(-10) = ` 1.0x10^-10 mol/L

adarshanurag's profile pic

adarshanurag | Student, Grade 11 | (Level 1) Valedictorian

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The solution is acidic. 

Reason pH is less than 7.

The ionic product of a solution = [H+] X [OH-] = 1 x 10^-14

                                            => 1 x 10^-4 X [OH-] = 1 x 10^-14   

                                            => [OH-] = 1 x 10^-10

adarshanurag's profile pic

adarshanurag | Student, Grade 11 | (Level 1) Valedictorian

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 pH of a solution is the negative reciprocal to the base ten of hydronium ion concentration

Given [H+] = 1 x 10^-4

pH is defined as -log [H+]

hence pH of solution = -log ( 10^-4)

                             = -(-4) log (10)

                             = 4

saikrishna1032's profile pic

saikrishna1032 | College Teacher | (Level 1) eNoter

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given [H+] = 10^-4` `

pH is defined as -log [H+]

hence pH of solution = -log ( 10^-4)

                             = -(-4) log (10)

                             = 4

solution is acidic , any aqueos solution whose pH < 7  is always acidic

in aqueos solutions  [H+]=[OH-] = 10^-14

hence [OH-] =(10 ^-14)/10^-4 

           [OH-] = 10^-10 .

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