Respuesta :
Answer
[H⁺] = 1.778 x 10⁻³ M
[OH⁻] = 5.624 x 10⁻¹² M
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
[tex]pH=-\log_[H^+][/tex]Putting the values of pH as 2.75 into the equation above, the [H⁺] of the juice can be calculated as follows:
[tex]\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\times10^{-3}M \end{gathered}[/tex]The [H⁺] = 1.778 x 10⁻³ M
The [H⁺] and [OH⁻] of an aqeuous solution is related by
[tex]\begin{gathered} [H^+]\times[OH^-]=1.0\times10^{-14} \\ \end{gathered}[/tex]So, putting [H⁺] = 1.778 x 10⁻³ M into the relation, we have [OH⁻] of the lemon juice to be:
[tex]\begin{gathered} 1.778\times10^{-3}\times[OH^-]=1.0\times10^{-14} \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }1.778\times10^{-3} \\ \\ \frac{1.778\times10^{-3}\times[OH^-]}{1.778\times10^{-3}}=\frac{1.0\times10^{-14}}{1.778\times10^{-3}} \\ \\ .[OH^-]=5.624\times10^{-12}M \end{gathered}[/tex]Therefore, [H⁺] = 1.778 x 10⁻³ M and [OH⁻] = 5.624 x 10⁻¹² M
