Question

1. complete the reaction mg(oh)₂ → mg⁺² + 2 1

there is complete ionization

1. what is the ph of a 0.001000 m mg(oh)₂ solution ?

first, find the oh⁻¹ using stoichiometry
2 mol oh⁻¹
(1.0000 x 10⁻³ m mg(oh)₂ )(-----------) = 3 m oh⁻¹
4 mol mg(oh)₂
then, find the h₃o⁺¹ using k_w
k_w 5
h₃o⁺¹ = ------- = ------- = 6
7 8 ph = -log( 9 ) = 10
a. acetic b. hydrochloric c. phosphoric d. perchloric e. hydrofluoric
f. sulfuric g. sulfurous h. hydrosulfuric i. oh⁻¹ j. h₃o⁺¹ k. so₄⁻²
l. mg⁺² m. cl⁻¹ n. hpo₃⁻² o. h₂po₃⁻¹ p. co₃⁻² q. hco₃⁻¹
r. h₂s s. hs⁻¹ t. s⁻² u. 0.2500 v. 0.1000 w. 0.02500

Explanation:

Step1: Complete the reaction

According to the law of conservation of mass and charge, when \(Mg(OH)_2\) ionizes completely (\(Mg(OH)_2
ightarrow Mg^{2 +}+2OH^{-}\)), so the blank 1 is \(OH^{-}\) (option I).

Step2: Calculate \([OH^{-}]\)

From the stoichiometry of the reaction \(Mg(OH)_2
ightarrow Mg^{2 +}+2OH^{-}\), the mole ratio of \(OH^{-}\) to \(Mg(OH)_2\) is \(2:1\).
\((1.0000\times 10^{-3}\text{ M }Mg(OH)_2)\times\frac{2\text{ mol }OH^{-}}{1\text{ mol }Mg(OH)_2}=2\times10^{-3}\text{ M }OH^{-}\), so blank 2 is \(2\), blank 4 is \(1\), blank 3 is \(2.000\times 10^{- 3}\)

Step3: Calculate \([H_3O^{+}]\)

We know that \(K_w=[H_3O^{+}][OH^{-}]=1.0\times 10^{-14}\) at \(25^{\circ}C\). So \([H_3O^{+}]=\frac{K_w}{[OH^{-}]}\), blank 5 is \(H_3O^{+}\), blank 7 is \(OH^{-}\), blank 8 is \(2.000\times 10^{-3}\)
\([H_3O^{+}]=\frac{1.0\times 10^{-14}}{2.000\times 10^{-3}} = 5.0\times 10^{-12}\) (blank 6)

Step4: Calculate pH

\(pH=-\log[H_3O^{+}]\), blank 9 is \(5.0\times 10^{-12}\)
\(pH =-\log(5.0\times 10^{-12})=11.30\) (blank 10)

Answer:

1. I. \(OH^{-}\)

2.

• Blank 2: \(2\)
• Blank 3: \(2.000\times 10^{-3}\)
• Blank 4: \(1\)
• Blank 5: \(H_3O^{+}\)
• Blank 6: \(5.0\times 10^{-12}\)
• Blank 7: \(OH^{-}\)
• Blank 8: \(2.000\times 10^{-3}\)
• Blank 9: \(5.0\times 10^{-12}\)
• Blank 10: \(11.30\)