For each strong base solution, determine [OH – ], [H 3 O + ], pH, and pOH. b. 0.0112 M Ba(OH) 2

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.16 - Acids and Bases

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Tro 4th Edition

Ch.16 - Acids and Bases

Problem 82b

Chapter 16, Problem 82b

For each strong base solution, determine [OH^–], [H₃O⁺], pH, and pOH. b. 0.0112 M Ba(OH)₂

Verified step by step guidance

Determine the concentration of hydroxide ions, [OH^-], from the dissociation of Ba(OH)_2. Since Ba(OH)_2 is a strong base, it dissociates completely in water. Each formula unit of Ba(OH)_2 produces two OH^- ions. Therefore, [OH^-] = 2 \times 0.0112 \, M.

Calculate the concentration of hydronium ions, [H_3O^+], using the relationship between [OH^-] and [H_3O^+]. The product of [OH^-] and [H_3O^+] is equal to the ion product of water, K_w = 1.0 \times 10^{-14} \, at \, 25^\circ C. Use the formula [H_3O^+] = \frac{K_w}{[OH^-]}.

Determine the pOH of the solution using the formula pOH = -\log[OH^-].

Calculate the pH of the solution using the relationship between pH and pOH: pH + pOH = 14.

Summarize the results: [OH^-], [H_3O^+], pH, and pOH for the 0.0112 M Ba(OH)_2 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 and alkaline earth metal hydroxides, such as Ba(OH)2. The concentration of the base directly correlates to the concentration of OH- ions in solution.

pH and pOH Calculations

pH and pOH are measures of the acidity and basicity of a solution, respectively. pH is calculated as the negative logarithm of the hydronium ion concentration ([H3O+]), while pOH is 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.

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 allows for the calculation of [H3O+] when [OH-] is known, and vice versa, which is essential for determining pH and pOH in strong base solutions.