1. If I placed 2.5 grams of Ca(OH)2 into 500.0 mL of deionized water... What would my poH be? (to
correct sig figs) *
(1 Point)

0.866

Explanation:

To solve this problem, first, we need to calculate the molarity of Ca(OH)2. This is shown below:

Molar Mass of Ca(OH)2 = 40 + 2(16+1) = 40 + 2(17) = 40 + 34 = 74g/mol

Mass of Ca(OH)2 = 2.5g

Number of mole of Ca(OH)2 =?

Number of mole = Mass /Molar Mass

Number of mole of Ca(OH)2 = 2.5/74 = 0.034 mole

Now we can obtain the molarity as shown below:

Volume = 500mL = 500/1000 = 0.5L

Mole = 0.034 mole

Molarity = mole/Volume

Molarity = 0.034/0.5

Molarity = 0.068M

Next, let us generate a balanced dissociation equation for Ca(OH)2. This is illustrated below:

Ca(OH)2 —> Ca^2+ + 2OH^-

From the equation above,

1 mole of Ca(OH)2 produced 2 moles of OH^-.

Therefore, 0.068M of Ca(OH)2 will produce = 0.068M x 2 = 0.136M of OH^-.

Now we can calculate the pOH as follow:

[OH^-] = 0.136M

pOH = — Log [OH^-]

pOH = — Log 0.136

pOH = 0.866

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