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3 years ago

What is the hydroxide [OH-] concentration of a solution that has a pOH of 4.90? 14 14 1.26 x10-5 1.26 x10, -5 9.1 9.1 7.94 x 104

Chemistry

1 answer:

polet [3.4K]3 years ago

3 0

Answer:

The hydroxide [OH-] concentration of the solution is 1.26*10⁻⁵ M.

Explanation:

The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution.

POH indicates the concentration of hydroxyl ions [OH-] present in a solution and is defined as the negative logarithm of the activity of hydroxide ions (that is, the concentration of OH- ions):

pOH= -log [OH-]

A solution has a pOH of 4.90. Replacing in the definition of pOH:

4.90= -log [OH-]

Solving:

-4.90= log [OH-]

1.26*10⁻⁵ M= [OH-]

The hydroxide [OH-] concentration of the solution is 1.26*10⁻⁵ M.

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Since we know that the sulfuric acid is the limiting reactant in this chemical reaction, we know that we are going to be left with excess aluminum hydroxide. So to find the amount of leftover aluminum hydroxide we are going to need to convert the given amount of sulfuric acid to the amount of aluminum hydroxide needed to react with the sulfuric acid.

$\frac{35g H_{2}SO_{4}}{1}*\frac{1 mole H_{2}SO_{4}}{98.079 g H_{2}SO_{4}} *\frac{2 moles Al(OH)_{3} }{3 moles H_{2}SO_{4}} * \frac{78.003 g Al(OH)_{3} }{1 mole Al(OH)_{3} } = 18.557 g Al(OH)_{3}$

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