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2 years ago

Calculate the hydronium ion concentration in an aqueous solution with a poh of 4.33 at 25°c.

1 answer:

4 0

Taking into account the definition of pH and pOH, the hydronium ion concentration in an aqueous solution with a pOH of 4.33 at 25°c is 2.138×10⁻¹⁰ M.

<h3>Definition of pH</h3>

pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.

The pH is defined as the negative base 10 logarithm of the activity of hydrogen ions, that is, the concentration of hydrogen ions or hydronium ion H₃O⁺:

pH= - log [H⁺]= - log [H₃O⁺]

<h3>Definition of pOH</h3>

Similarly, pOH is a measure of hydroxyl ions in a solution and is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

<h3>Relationship between pH and pOH</h3>

The following relationship can be established between pH and pOH:

pOH + pH= 14

<h3>Concentration of hydronium ions</h3>

Being pOH= 4.33, pH is calculated as:

pH + 4.33= 14

pH= 14 - 4.33

Replacing in the definition of pH the concentration of hydronium ions is obtained:

- log [H₃O⁺]= 9.67

Solving:

[H₃O⁺]= 10⁻⁹ ⁶⁷

Finally, the hydronium ion concentration in an aqueous solution with a pOH of 4.33 at 25°c is 2.138×10⁻¹⁰ M.

Learn more about pH and pOH:

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% composition = (mass of element / total mass) * 100

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Oxalic Acid, a compound found in plants and vegetables such as rhubarb, has a mass percent composition of 26.7% C, 2.24% H, and

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From the information given:

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To determine the empirical formula;

First thing is to find the numbers of moles of each atom.

For Carbon:

$=26.7 \ g\times \dfrac{1 \ mol }{12.01 \ g} \\ \\ =2.22 \ mol \ of \ Carbon$

For Hydrogen:

$=2.24 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =2.22 \ mol \ of \ Hydrogen$

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$=71.1 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =4.44 \ mol \ of \ oxygen$

Now; we use the smallest no of moles to divide the respective moles from above.

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$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ carbon$

For Hydrogen:

$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ hydrogen$

For Oxygen:

$\dfrac{4.44 \ mol \ of \ Oxygen}{2.22} =2 \ mol \ of \ oxygen$

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