Question

Lemon juice, pH = 2.75 Calculate [H +] and [OH﹘]

Answer 1

[H⁺] = 1.778 x 10⁻³ M

[OH⁻] = 5.624 x 10⁻¹² M

Step-by-step explanation

Given:

The pH of the lemon juice = 2.75

What to find:

The [H⁺] and [OH⁻] of the lemon juice.

Step-by-step solution:

pH is a measure of hydrogen ion concentration, a measure of the acidity or alkalinity of a solution. The pH scale usually ranges from 0 to 14.

The equation for calculating pH is

`pH=-\log_[H^+]`

Putting the values of pH as 2.75 into the equation above, the [H⁺] of the juice can be calculated as follows:

`\begin{gathered} 2.75=-\log_[H^+] \\ \\ Multiply\text{ }all\text{ }through\text{ }by\text{ }- \\ \\ \log_[H^+]=-2.75 \\ \\ .[H^+]=10^(-2.75) \\ \\ .[H^+]=1.778*10^(-3)M \end{gathered}`

The [H⁺] = 1.778 x 10⁻³ M

The [H⁺] and [OH⁻] of an aqeuous solution is related by

`\begin{gathered} [H^+]*[OH^-]=1.0*10^(-14) \\ \end{gathered}`

So, putting [H⁺] = 1.778 x 10⁻³ M into the relation, we have [OH⁻] of the lemon juice to be:

`\begin{gathered} 1.778*10^(-3)*[OH^-]=1.0*10^(-14) \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }1.778*10^(-3) \\ \\ (1.778*10^(-3)*[OH^-])/(1.778*10^(-3))=(1.0*10^(-14))/(1.778*10^(-3)) \\ \\ .[OH^-]=5.624*10^(-12)M \end{gathered}`

Therefore, [H⁺] = 1.778 x 10⁻³ M and [OH⁻] = 5.624 x 10⁻¹² M