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What is the concentration of H+ ions at a pH = 8?

skelet666 [1.2K]

Taking into account the definition of pH and pOH, the concentration of H⁺ ions at pH=8 is 1×10⁻⁸ M and the concentration of OH⁻ ions at pH=8 is 1×10⁻⁶ M.

<h3>Definition of pH</h3>

pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.

The pH is defined as the negative base 10 logarithm of the activity of hydrogen ions, that is, the concentration of hydrogen ions or H₃O⁺:

pH= - log [H⁺]= - log [H₃O⁺]

<h3>Definition of pOH</h3>

Similarly, pOH is a measure of hydroxyl ions in a solution and is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

<h3 /><h3>Relationship between pH and pOH</h3>

The following relationship can be established between pH and pOH:

pOH + pH= 14

<h3>Concentration of H⁺ ions</h3>

In this case, pH=8. Replacing in the definition of pH:

8= - log [H⁺]

Solving:

Finally, the concentration of H⁺ ions at pH=8 is 1×10⁻⁸ M.

<h3>Concentration of OH⁻ ions</h3>

Being pH= 8, pOH is calculated as:

pOH + 8= 14

pOH= 14 - 8

pOH= 6

Replacing in the definition of pOH the concentration of OH⁻ ions is obtained:

- log [OH⁻]= 6

Solving

Finally, the concentration of OH⁻ ions at pH=8 is 1×10⁻⁶ M.

Learn more about pH and pOH:

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7 0

2 years ago

How can l calculate the no of constraints?

Kisachek [45]

Here u go I will submit other help for u

4 0

3 years ago

Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a ho

antiseptic1488 [7]

To solve this problem it is necessary to apply the concepts related to Malus' law. Malus' law indicates that the intensity of a linearly polarized ray of light that passes through a perfect analyzer with a vertical optical axis is equivalent to:

$I = I_0 cos^2(\theta)$

$I_0 =$ Indicates the intensity of the light before passing through the Polarizer,

I = The resulting intensity, and

$\theta$ = Indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

There is 3 polarizer, then

For the exit of the first polarizer we have that the intensity is,

$I_2 = I_0cos^2(45)$

For the third polarizer then we have,

$I_3 = I_2 cos^{2}(45)$

Replacing with the first equation,

$I_3 = I_0cos^2(45)cos^{2}(45)$

$I_3 = I_0 (\frac{1}{2})(\frac{1}{2})$

$I_3 = I_0 \frac{1}{4}$

Therefore the transmitted intensity now is $\frac{1}{4}$ of the initial intensity.

5 0

4 years ago

The absolute temperature of a gas is t. in order to double the rms speed of its molecules, what should be the new absloute tempe

jenyasd209 [6]

The new absloute temperature should be 4t.

<h3>Temperature </h3>

The hotness of matter or radiation is expressed by the physical quantity known as temperature.

There are three different types of temperature scales: those, like the SI scale, that are defined in terms of the average translational kinetic energy per freely moving microscopic particle, like an atom, molecule, or electron in a body; those that solely depend on strictly macroscopic properties and thermodynamic principles, like Kelvin's original definition; and those that are not defined by theoretical principles but rather by useful empirical properties of particula.

Using a thermometer, one can gauge temperature. It is calibrated using different temperature scales, each of which historically defined itself using a different set of reference points and thermometric materials.

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6 0

2 years ago

Charge is uniformly distributed throughout a spherical insulating volume of radius R=4.00 cm . The charge per unit volume is −6.

aleksley [76]

Answer

-8.67× 10^6 N/C

Explanation:

The Electric Field is defined as force per unit charge.

E = Q/ 4π£r2

Qv= −6.5 μCm3

Qv = Q/ V= Q/ 4/3 πr3

Hence Q = 4/3 πr3 × Qv

Hence E = 4/3 πr3 × Qv / 4π£r2= Qvr/3£

−6.5 μ × 4/ 3×8.854 ×10^-12

-6.5 × 4 × 10^6/3 = -8.67× 10^6 N/C

Note: £ = 8.854×10^-12m/F

is the permittivity of free space

Qv is the charge per unit volume

V is volume and volume

8 0

3 years ago

What are the benefits that you can get from playing volleyball​

What are the benefits that you can get from playing volleyball