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3 years ago

8

Is the sun typical of these nearby stars

1 answer:

marshall27 [118]3 years ago

8 0

No the sun is the biggest star in the universe.

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Acid & Base Calculations Calculate the hydronium ion concentration for each. Tell whether it is an acid or a base. 1. pH = 5

Anuta_ua [19.1K]

<u>Explanation:</u>

pH is the negative logarithm of hydronium ion concentration present in a solution.

If the solution has high hydrogen ion concentration, then the pH will be low and the solution will be acidic. The pH range of acidic solution is 0 to 6.9
If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14
The solution having pH equal to 7 is termed as neutral solution.

To calculate the pH of the solution, we use equation:

$pH=-\log[H_3O^+]$ ......(1)

To calculate the pOH of the solution, we use the equation:

pH + pOH = 14 ........(2)

<u>For 1:</u>

We are given:

pH = 5.54

Putting values in equation 1, we get:

$5.54=-\log[H_3O^+]$

$[H_3O^+]=2.88\times 10^{-6}M$

Now, putting values in equation 2, we get:

14 = 5.54 + pOH

pOH = 8.46

The solution is acidic in nature.

<u>For 2:</u>

We are given:

pOH = 9.7

Putting values in equation 2, we get:

14 = 9.7 + pH

pH = 4.3

Now, putting values in equation 1, we get:

$4.3=-\log[H_3O^+]$

$[H_3O^+]=5.012\times 10^{-5}M$

The solution is acidic in nature.

<u>For 3:</u>

We are given:

pH = 7.0

Putting values in equation 1, we get:

$7.0=-\log[H_3O^+]$

$[H_3O^+]=1.00\times 10^{-7}M$

Now, putting values in equation 2, we get:

14 = 7.0 + pOH

pOH = 7.0

The solution is neither acidic nor basic in nature.

<u>For 4:</u>

We are given:

pH = 12.9

Putting values in equation 1, we get:

$12.9=-\log[H_3O^+]$

$[H_3O^+]=1.26\times 10^{-13}M$

Now, putting values in equation 2, we get:

14 = 12.9 + pOH

pOH = 1.1

The solution is basic in nature.

<u>For 5:</u>

We are given:

pOH = 1.2

Putting values in equation 2, we get:

14 = 1.2 + pH

pH = 12.8

Now, putting values in equation 1, we get:

$12.8=-\log[H_3O^+]$

$[H_3O^+]=1.58\times 10^{-13}M$

The solution is basic in nature.

<u>For 6:</u>

We are given:

$[H_3O^+]=1\times 10^{-5}M$

Putting values in equation 1, we get:

$pH=-\log(1\times 10^{-5})$

$pH=5$

Now, putting values in equation 2, we get:

14 = 5 + pOH

pOH = 9

The solution is acidic in nature.

5 0

3 years ago