Question

1. complete the reaction naoh → na⁺¹ + 1

there is complete ionization

1. what is the ph of a 0.001000 m naoh solution ?

first, find the h₃o⁺¹ using k_w
k_w 2
h₃o⁺¹ = --------- = --------- = 3 m
4 5 ph = -log(6) = 7
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
x. 1.000×10⁻³ y. 3.0000 z. 1×10⁻¹⁴ aa. 2.6990 bb. 1.000×10⁻¹¹

cc. 11.0000 dd. 2 ee. 1 ff. 2.000×10⁻³ gg. 5.000×10⁻¹²
hh. 11.3010 ii. 0.01250 jj. 0.22100 kk. 0.05656 ll. acid
mm. conjugate acid oo. base pp. conjugate base

Explanation:

Step1: Complete the ionization reaction of NaOH

NaOH is a strong base. In aqueous solution, it completely ionizes into \(Na^{+}\) and \(OH^{-}\). So, the blank 1 is \(I\) (\(OH^{-1}\)).

Step2: Recall the formula for \(K_{W}\)

The ion - product constant of water is \(K_{W}=[H_{3}O^{+}][OH^{-}]\). Given a \(0.001000\ M\ NaOH\) solution, \([OH^{-}]=0.001000\ M\) (from the ionization of \(NaOH\)). So, blank 2 is \(OH^{-1}\), blank 4 is \(OH^{-1}\), and blank 5 is \(H_{3}O^{+1}\).

Step3: Calculate \([H_{3}O^{+}]\)

We know that \(K_{W} = 1.0\times10^{-14}\) at \(25^{\circ}C\). Using the formula \([H_{3}O^{+}]=\frac{K_{W}}{[OH^{-}]}\), substituting \(K_{W}=1.0\times 10^{-14}\) and \([OH^{-}]=1.000\times10^{-3}\ M\), we get \([H_{3}O^{+}]=\frac{1.0\times10^{-14}}{1.000\times10^{-3}}=1.000\times10^{-11}\ M\). So, blank 3 is \(BB\) (\(1.000\times10^{-11}\)), blank 6 is \(H_{3}O^{+1}\) (or the value \(1.000\times10^{-11}\)).

Step4: Calculate the \(pH\)

The formula for \(pH\) is \(pH =-\log[H_{3}O^{+}]\). Substituting \([H_{3}O^{+}]=1.000\times10^{-11}\), we get \(pH=-\log(1.000\times 10^{-11}) = 11.0000\). So, blank 7 is \(CC\) (\(11.0000\)).

Answer:

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

2.

• Blank 2: \(I\) (\(OH^{-1}\))
• Blank 3: \(BB\) (\(1.000\times10^{-11}\))
• Blank 4: \(I\) (\(OH^{-1}\))
• Blank 5: \(IJ\) (\(H_{3}O^{+1}\))
• Blank 6: \(IJ\) (\(H_{3}O^{+1}\)) (or \(BB\) (\(1.000\times10^{-11}\)))
• Blank 7: \(CC\) (\(11.0000\))