Calculate the [H 3 O + ] of each aqueous solution with the following [OH - ]: d. bile, 2.5 × 10 -6 M

Hello. And this primer has to find the hydro concentration of the human aqueous humor. If its hydroxyline concentration is 2.2 times to the minus seven molar, we can find the hydro concentration by making use of the iron product cost of water which tells us that the hydro ion concentration times the hydroxide ion concentration is equal to one times 10 to minus 14 at 25 degrees Celsius hydro concentration. Then it is equal to one times 10 to minus 14 by, by the hydroxid concentration. This is equal to one times 10 minus 14, divided by 2.2 times 10 to minus seven, which is equal to 4.5 times 10 to minus eight moller. So the hydro concentration of human aqueous humor is 4.5 times to the minus eight molar. Thanks for watching. Hope this helped.

Calculate the [H3O+] of each aqueous solution with the following [OH-]:d. bile, 2.5 × 10-6 M

Step 1: Recall the relationship between [H₃O⁺] and [OH⁻] in aqueous solutions, which is governed by the ion-product constant for water (Kw). At 25°C, Kw = 1.0 × 10⁻¹⁴. The formula is: K w = [H 3 O ] × [OH - . Step 2: Rearrange the formula to solve for [H₃O⁺]: [H 3 O ] = K w [OH - . Step 3: Substitute the given value of [OH⁻] (2.5 × 10⁻⁶ M) and Kw (1.0 × 10⁻¹⁴) into the formula: [H 3 O ] = 1.0×10 -14 2.5×10 -6 . Step 4: Perform the division to calculate [H₃O⁺]. Ensure you handle the exponents correctly by subtracting the exponents in the numerator and denominator. Step 5: Express the final result in proper scientific notation, ensuring the significant figures match the precision of the given data (2 significant figures).

Ch.10 Acids and Bases and Equilibrium

Timberlake - Chemistry: An Introduction to General, Organic, and Biological Chemistry 13th Edition

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Timberlake 13th Edition

Ch.10 Acids and Bases and Equilibrium

Problem 33d

Chapter 10, Problem 33d

Calculate the [H₃O⁺] of each aqueous solution with the following [OH^-]:
d. bile, 2.5 × 10^-6 M

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Step 1: Recall the relationship between [H₃O⁺] and [OH⁻] in aqueous solutions, which is governed by the ion-product constant for water (Kw). At 25°C, Kw = 1.0 × 10⁻¹⁴. The formula is:

K_{w} = {[H}_{3 O}] \times {[OH}_{-}

Step 2: Rearrange the formula to solve for [H₃O⁺]:

{[H}_{3 O}] = \frac{K_{w}}{{[OH}_{-}}

Step 3: Substitute the given value of [OH⁻] (2.5 × 10⁻⁶ M) and Kw (1.0 × 10⁻¹⁴) into the formula:

{[H}_{3 O}] = \frac{1.0×10-14}{2.5×10-6}

Step 4: Perform the division to calculate [H₃O⁺]. Ensure you handle the exponents correctly by subtracting the exponents in the numerator and denominator.

Step 5: Express the final result in proper scientific notation, ensuring the significant figures match the precision of the given data (2 significant figures).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

pH and pOH Relationship

The pH and pOH of a solution are related through the equation pH + pOH = 14 at 25°C. This relationship allows us to calculate the concentration of hydronium ions [H₃O⁺] from the concentration of hydroxide ions [OH⁻]. By knowing the [OH⁻], we can find pOH, and subsequently derive pH, which gives us [H₃O⁺] using the formula [H₃O⁺] = 10^(-pH).

Ion Product of Water (Kw)

The ion product of water (Kw) is a constant at a given temperature, defined as Kw = [H₃O⁺][OH⁻]. At 25°C, Kw is equal to 1.0 x 10⁻¹⁴. This relationship is crucial for calculating the concentration of hydronium ions when the concentration of hydroxide ions is known, as it allows us to rearrange the equation to find [H₃O⁺] = Kw / [OH⁻].

Concentration Units in Chemistry

Concentration in chemistry is often expressed in molarity (M), which is moles of solute per liter of solution. Understanding how to manipulate these units is essential for calculations involving [H₃O⁺] and [OH⁻]. In this context, the given concentration of bile as 2.5 x 10⁻⁶ M [OH⁻] must be used to derive the corresponding [H₃O⁺] concentration through the established relationships.