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2 years ago

What is the pH of 3.65 grams of HCl dissolved in 1250 mL distilled H2O

Chemistry

1 answer:

slavikrds [6]2 years ago

6 0

Considering the definition of pH, the pH of 3.65 grams of HCl dissolved in 1250 mL distilled H₂O is 1.0969.

<h3>Definition of pH</h3>

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

Mathematically, pH is calculated as the negative base 10 logarithm of the activity of hydrogen ions:

pH= - log [H⁺]

The numerical scale that measures the pH of substances includes the numbers from 0 to 14. The pH value 7 corresponds to neutral substances. Acidic substances are those with a pH lower than 7, while basic substances have a pH higher than 7.

<h3>Concentration of HCl</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

$Molarity=\frac{number of moles}{volume}$

In this case, you know:

number of moles= $3.65 grams x\frac{1 mole}{36.45 grams} =$ 0.1 moles (being 36.45 g/mole the molar mass of HCl)
volume= 1250 mL= 1.250 L (being 1000 mL= 1 L)

Replacing in the definition of molarity:

$Molarity=\frac{0.1 moles}{1.250 L}$

Solving:

Molarity= 0.08 $\frac{moles}{liter}$ = 0.08 M

<h3>pH of the solution</h3>

Hydrochloric acid is a strong acid, so in aqueous solution it will be totally dissociated. So the concentration of protons is equal to the initial concentration of acid. This is: [H⁺]= [HCl]= 0.08 M

Replacing in the definition of pH:

pH= - log (0.08 M)

Solving:

pH= 1.0969

Finally, the pH of 3.65 grams of HCl dissolved in 1250 mL distilled H₂O is 1.0969.

Learn more about

pH:

brainly.com/question/3992824

brainly.com/question/12200689

molarity:

brainly.com/question/9324116

brainly.com/question/10608366

brainly.com/question/7429224

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Explanation:

Hi there!!

you asked to multiply these all right,

you can simply multiply it ,

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