Question

Complete the following table. SHOW YOUR WORK in the space underneath the table. [H3O+] pH [OH-] pOH Acidic/basic/ neutral? 3.5 x 10-6 12.45 9.6 x 10-10 6.17

Answer 1

To complete the table, we need to understand the relationships between [H3O+], pH, [OH-], and pOH. These relationships are defined by the following equations:

`\text{pH} = -\log [H3O^+]`

`\text{[H3O^+][OH^-]} = 1 * 10^(-14) \ \text{(at 25°C)}`

`\text{pOH} = -\log [OH^-]`

`\text{pH} + \text{pOH} = 14`

Given [H3O+] = 3.5 x 10^-6:

`\text{pH} = -\log (3.5 * 10^(-6))`

Calculating the pH:

`\text{pH} \approx -(-5.46)`

`\text{pH} \approx 5.46`

(This corrects the pH value provided in your table, which should be close to our calculated value for consistency within the relationships outlined above.)

Given [OH-] = 9.6 x 10^-10:

`\text{pOH} = -\log (9.6 * 10^(-10))`

Calculating the pOH:

`\text{pOH} \approx -(-9.02)`

`\text{pOH} \approx 9.02`

(This corrects the pOH value provided in your table, which should be close to our calculated value for consistency.)

To determine whether the solution is acidic, basic, or neutral:

- Acidic if pH < 7 (or equivalently if pOH > 7)

- Basic if pH > 7 (or equivalently if pOH < 7)

- Neutral if pH = 7 (or equivalently if pOH = 7)

Since the pH is approximately 5.46, which is less than 7, the solution is acidic.

Now let's correct the table using the calculated values derived from the relationships:

`[H3O^+] = 3.5 * 10^(-6)`

`\text{pH} \approx 5.46`

`[OH^-] = (1 * 10^(-14))/(3.5 * 10^(-6))`

Calculating [OH-]:

`[OH^-] \approx 2.86 * 10^(-9)`

`\text{pOH} \approx 9.02`