For each strong base solution, determine [OH – ], [H 3 O + ], pH, and pOH. a. 8.77×10 –3 M LiOH

hey everyone we're asked to solve for our P. O. H. Ph concentration of hydro ni um ions and concentration of hydroxide ions In a 1.2 times 10 to the negative 3rd moller strontium hydroxide strong based solution First. We can go ahead and solve our concentration of hydroxide ions by taking our 1.2 Times 10 to the negative 3rd molar of strontium hydroxide. And we can see through our multiple ratios that one mole of strontium hydroxide contains two mole of hydroxide ions, solving for our hydroxide ion concentration. We end up with a value of 2.4 times 10 to the negative third molar of hydroxide ions, solving for our P. O. H. We can then go ahead And take the negative log of 2. Times 10 to the -3 Molar. This will get us to a p. h. of 2.62, Solving for our ph we can then take 14 and subtract 2.62 which is our P. O. H. This will get us to a ph of 11.38. Now, lastly to solve for our concentration of hydro ni um ions We can take 10 to the negative 11. And this will get us to a concentration of 4.2 Times 10 to the -12 Molar and this is going to be our final answers for this question. So I hope this made sense. And let us know if you have any questions

Ch.17 - Acids and Bases

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Tro 5th Edition

Ch.17 - Acids and Bases

Problem 82a

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Chapter 17, Problem 82a

For each strong base solution, determine [OH^–], [H₃O⁺], pH, and pOH. a. 8.77×10^–3 M LiOH

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Step 1: Identify the strong base and its dissociation. LiOH is a strong base, which means it dissociates completely in water. The dissociation can be represented as: \[ \text{LiOH} \rightarrow \text{Li}^+ + \text{OH}^- \]. Therefore, the concentration of \( \text{OH}^- \) ions is equal to the concentration of the LiOH solution, which is \( 8.77 \times 10^{-3} \text{ M} \).

Step 2: Calculate the \( [\text{OH}^-] \). Since LiOH dissociates completely, \( [\text{OH}^-] = 8.77 \times 10^{-3} \text{ M} \).

Step 3: Use the ion product of water to find \( [\text{H}_3\text{O}^+] \). The ion product of water at 25°C is \( K_w = 1.0 \times 10^{-14} \). Use the formula \( [\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]} \) to find \( [\text{H}_3\text{O}^+] \).

Step 4: Calculate the pOH. Use the formula \( \text{pOH} = -\log[\text{OH}^-] \) to find the pOH of the solution.

Step 5: Calculate the pH. Use the relationship \( \text{pH} + \text{pOH} = 14 \) to find the pH of the solution.

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Key Concepts

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

Strong Bases

Strong bases are substances that completely dissociate in water to produce hydroxide ions (OH-). Common examples include alkali metal hydroxides like lithium hydroxide (LiOH). The concentration of the base directly correlates to the concentration of OH- ions in solution, which is crucial for calculating pH and pOH.

pH and pOH Calculations

pH is a measure of the acidity of a solution, defined as the negative logarithm of the hydrogen ion concentration ([H3O+]). Conversely, pOH measures the basicity, defined as the negative logarithm of the hydroxide ion concentration ([OH-]). The relationship between pH and pOH is given by the equation pH + pOH = 14 at 25°C, allowing for the conversion between these two measures.

Ion Product of Water

The ion product of water (Kw) is the equilibrium constant for the self-ionization of water, defined as Kw = [H3O+][OH-] = 1.0 x 10^-14 at 25°C. This relationship is essential for determining the concentrations of H3O+ and OH- in a solution, especially when dealing with strong bases, as it helps to find the corresponding [H3O+] when [OH-] is known.

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For each strong base solution, determine [OH–], [H3O+], pH, and pOH. a. 8.77×10–3 M LiOH

Verified Solution

Key Concepts

Strong Bases

pH and pOH Calculations

Ion Product of Water

For each strong base solution, determine [OH^–], [H₃O⁺], pH, and pOH. a. 8.77×10^–3 M LiOH