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 $\ce{h3o^{+1}} \ce{oh^{-1}} = 10^{-14}$
$10^{-14}$ $10^{-14}$
$\ce{h3o^{+1}} = \frac{------}{\ce{oh^{-1}}} = \frac{---------}{5.555 \times 10^{-4}} = __5__$

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

Explanation:

Step1: Solve for pH when $[\text{H}_3\text{O}^+] = 5.555 \times 10^{-4}$

The formula for pH is $\text{pH} = -\log[\text{H}_3\text{O}^+]$. Plugging in the value: $\text{pH} = -\log(5.555 \times 10^{-4})$. Calculating this, we get $\text{pH} \approx 3.255$ (which is option F). Since pH < 7, the solution is acidic (option A).

Step2: Solve for pOH when $[\text{OH}^-] = 5.555 \times 10^{-4}$

The formula for pOH is $\text{pOH} = -\log[\text{OH}^-]$. Plugging in the value: $\text{pOH} = -\log(5.555 \times 10^{-4}) \approx 3.255$ (option F). Since pOH < 7, pH = 14 - pOH = 10.745 > 7, so the solution is basic (option B).

Step3: Solve for $[\text{H}_3\text{O}^+]$ when $[\text{OH}^-] = 5.555 \times 10^{-4}$

Using $[\text{H}_3\text{O}^+][\text{OH}^-] = 10^{-14}$, we get $[\text{H}_3\text{O}^+] = \frac{10^{-14}}{5.555 \times 10^{-4}} \approx 1.800 \times 10^{-11}$ (option D).

Step4: Solve for pH using $[\text{H}_3\text{O}^+] = 1.800 \times 10^{-11}$

$\text{pH} = -\log(1.800 \times 10^{-11}) \approx 10.745$ (option E). Since pH > 7, the solution is basic (option B).

Answer:

1. pH: F. 3.255; Solution type: A. acidic
2. pOH: F. 3.255; Solution type: B. basic or alkaline
3. $[\text{H}_3\text{O}^+]$: D. $1.800 \times 10^{-11}$; pH: E. 10.745; Solution type: B. basic or alkaline