Question

use your calculator for the following

1. when $\ce{h3o^{+1}} = 5.555 \times 10^{-4}$, the ph = 1

this is an 2 solution (acidic or basic)

1. when $\ce{oh^{-1}} = 5.555 \times 10^{-4}$, the poh = 3

this is an 4 solution (acidic or basic)

1. when $\ce{oh^{-1}} = 5.555 \times 10^{-4}$, what is the ph ?

using $\ce{h3o^{+1}}\ce{oh^{-1}} = 10^{-14}$
\qquad\qquad $10^{-14}$ \qquad\quad $10^{-14}$
$\ce{h3o^{+1}} = \frac{\quad\quad\quad}{\ce{oh^{-1}}} = \frac{\quad\quad\quad\quad\quad}{5.555 \times 10^{-4}} = 5

ph = -log6 = 7
this is an 8 solution. (acidic or basic)
a. acidic \quad b. basic or alkaline \quad c. neutral \quad d. $1.800 \times 10^{-11}$
e. 10.745 \quad f. 3.255

Explanation:

Step1: Calculate pH for [H₃O⁺] = 5.555×10⁻⁴

pH = -log[H₃O⁺] = -log(5.555×10⁻⁴) ≈ 3.255 (F)
Since pH < 7, solution is acidic (A).

Step2: Calculate pOH for [OH⁻] = 5.555×10⁻⁴

pOH = -log[OH⁻] = -log(5.555×10⁻⁴) ≈ 3.255 (F)
Since pOH < 7, [OH⁻] < [H₃O⁺]? Wait, no: pOH = 3.255 means pH = 14 - 3.255 = 10.745 > 7, so basic (B). Wait, step2: pOH is 3.255 (F), and since pOH < 7, [OH⁻] is 5.555×10⁻⁴, so [H₃O⁺] = 10⁻¹⁴ / 5.555×10⁻⁴ ≈ 1.800×10⁻¹¹ (D), pH = 10.745 (E) > 7, so basic (B). Wait, the second question: when [OH⁻] = 5.555×10⁻⁴, pOH = -log(5.555×10⁻⁴) ≈ 3.255 (F). Then, since pOH < 7, the solution has [OH⁻] = 5.555×10⁻⁴, [H₃O⁺] = 10⁻¹⁴ / 5.555×10⁻⁴ ≈ 1.800×10⁻¹¹, so pH = 10.745 > 7, so basic (B). Wait, the second question's part 4: solution is basic (B).

Step3: Solve part 3 (when [OH⁻] = 5.555×10⁻⁴, find pH)

First, [H₃O⁺] = 10⁻¹⁴ / [OH⁻] = 10⁻¹⁴ / (5.555×10⁻⁴) ≈ 1.800×10⁻¹¹ (D, _5_). Then pH = -log[H₃O⁺] = -log(1.800×10⁻¹¹) ≈ 10.745 (E, _7_). Since pH > 7, solution is basic (B, _8_).

Now, let's list the answers:

1. pH = -log(5.555×10⁻⁴) ≈ 3.255 (F)
2. Solution is acidic? Wait no: first question: [H₃O⁺] = 5.555×10⁻⁴, pH = 3.255 < 7, so acidic (A, _2_).

Second question: [OH⁻] = 5.555×10⁻⁴, pOH = 3.255 (F, _3_), solution is basic (B, _4_).

Third question:

_5_: [H₃O⁺] = 10⁻¹⁴ / 5.555×10⁻⁴ ≈ 1.800×10⁻¹¹ (D)

_6_: [H₃O⁺] is 1.800×10⁻¹¹ (D), so pH = -log(1.800×10⁻¹¹) ≈ 10.745 (E, _7_), solution is basic (B, _8_).

Now, let's structure each part:

1. pH = -log(5.555×10⁻⁴) ≈ 3.255 (F)

Solution is acidic (A)

1. pOH = -log(5.555×10⁻⁴) ≈ 3.255 (F)

Solution is basic (B) (since pH = 14 - 3.255 = 10.745 > 7)
3.
_5_: [H₃O⁺] = 10⁻¹⁴ / 5.555×10⁻⁴ ≈ 1.800×10⁻¹¹ (D)
_6_: [H₃O⁺] (D)
_7_: pH = -log(1.800×10⁻¹¹) ≈ 10.745 (E)
_8_: basic (B)

Now, let's answer each blank:

1. 1: F (3.255)
2. 2: A (acidic)
3. 3: F (3.255)
4. 4: B (basic or alkaline)
5. 5: D (1.800×10⁻¹¹)
6. 6: D (1.800×10⁻¹¹) [or the [H₃O⁺] value, which is D]
7. 7: E (10.745)
8. 8: B (basic or alkaline)

Answer:

s (filling blanks):

1. F. 3.255
2. A. acidic
3. F. 3.255
4. B. basic or alkaline
5. D. 1.800 x 10⁻¹¹
6. D. 1.800 x 10⁻¹¹
7. E. 10.745
8. B. basic or alkaline