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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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slava [35]

<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of iron(III) phosphate = 20.00 g

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Putting values in equation 1, we get:

$\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol$

The given chemical equation follows:

$2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4$

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = $\frac{1}{2}\times 0.133=0.0665moles$ of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

$0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g$

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

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