Question

The pH of milk, black coffee, tomato juice, lemon juice, and egg white are 6.8,5.0, 4.2,2.2, and 7.8 respectively. Calculate the corresponding hydrogen ion concentration in each.

Answer 1

pH is a scale to measure the acidity and basicity of the aqueous solution. The term pH stands for the potential of hydrogen or as the power of hydrogen. The pH is a logarithmic value and it is inversely proportional to the concentration of hydrogen ions in the solution.

Complete step by step answer:
The hydrogen ion ${{\text{H}}^{\text{+}}}$ concentration in the solution is related to the pH of the solution. The pH and concentration ${{\text{H}}^{\text{+}}}$are related as,
$\text{pH=}\dfrac{\text{1}}{\text{log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ }}\text{=-log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ }$
Where, $\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ = concentration of hydrogen ion}$

Here, we have to find the concentration of ${{\text{H}}^{\text{+}}}$an ion in given examples.
i)For milk,
We are provided with a pH of milk as 6.8
Since we know that,
$\begin{aligned} & \text{pH=-log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ & \text{6}\text{.8=-log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ & \text{-6}\text{.8=log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ \end{aligned}$
Take an antilog on both sides,
$\begin{aligned} & \text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ =1}{{\text{0}}^{\text{-6}\text{.8}}} \\\ & \text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ =1}\text{.58}\times \text{1}{{\text{0}}^{\text{-7}}}\text{M} \\\ \end{aligned}$
The concentration of ${{\text{H}}^{\text{+}}}$ milk is equal to $\text{1}\text{.58}\times \text{1}{{\text{0}}^{\text{-7}}}\text{M}$.

iI)For black coffee,
We are provided with a pH of black coffee as 5.0
Since we know that ,
$\begin{aligned} & \text{pH=-log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ & \text{5}\text{.0=-log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ & \text{-5}\text{.0=log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ \end{aligned}$
Take an antilog on both sides,
$\begin{aligned} & \text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ =1}{{\text{0}}^{\text{-5}\text{.0}}} \\\ & \text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ =1}\text{.00}\times \text{1}{{\text{0}}^{-5}}\text{M} \\\ \end{aligned}$
The concentration of ${{\text{H}}^{\text{+}}}$ milk is equal to $\text{1}\text{.00}\times \text{1}{{\text{0}}^{\text{-5}}}\text{M}$.

iii)For tomato juice,
We are provided with a pH of tomato juice as 4.2
Since we know that ,
$\begin{aligned} & \text{pH=-log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ & \text{4}\text{.2=-log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ & \text{-4}\text{.2=log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ \end{aligned}$
Take an antilog on both sides,
$\begin{aligned} & \text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ =1}{{\text{0}}^{\text{-4}\text{.2}}} \\\ & \text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ =6}\text{.30}\times \text{1}{{\text{0}}^{-5}}\text{M} \\\ \end{aligned}$
The concentration of ${{\text{H}}^{\text{+}}}$ milk is equal to $\text{6}\text{.30}\times \text{1}{{\text{0}}^{\text{-5}}}\text{M}$.

iv)For the lemon juice,
We are provided with a pH of lemon juice as 2.2
Since we know that
$\begin{aligned} & \text{pH=-log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ & \text{2}\text{.2=-log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ & \text{-2}\text{.2=log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ \end{aligned}$
Take an antilog on both sides,
$\begin{aligned} & \text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ =1}{{\text{0}}^{\text{-2}\text{.2}}} \\\ & \text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ =6}\text{.30}\times \text{1}{{\text{0}}^{-3}}\text{M} \\\ \end{aligned}$
The concentration of ${{\text{H}}^{\text{+}}}$ milk is equal to $\text{6}\text{.30}\times \text{1}{{\text{0}}^{\text{-3}}}\text{M}$.

iv)For egg whites,
We are provided with the pH of egg whites as 7.8
Since we know that
$\begin{aligned} & \text{pH=-log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ & \text{7}\text{.8=-log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ & \text{-7}\text{.8=log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ \end{aligned}$
Take an antilog on both sides,
$\begin{aligned} & \text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ =1}{{\text{0}}^{\text{-7}\text{.8}}} \\\ & \text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ =1}\text{.58}\times \text{1}{{\text{0}}^{-8}}\text{M} \\\ \end{aligned}$
The concentration of ${{\text{H}}^{\text{+}}}$ milk is equal to$1.58\times \text{1}{{\text{0}}^{\text{-8}}}\text{M}$.

Additional Information:
The pH scale is used to measure the acidity and basicity of the solution. The scale has ranged from 0 to 14.
If $$$$$$\text{pH }\langle \text{ 7}$, the solution is acidic If$\text{pH }\rangle \text{ 7}$, the solution is basic If$\text{pH=7}$$,the solution is neutral

Note: Students can face a problem in solving the logarithmic values and taking the antilog of value. Remember that pH is always logarithmic of ${{\text{H}}^{\text{+}}}$ concentration with base 10. When you are taking antilog then take the value of pH given as the negative power of 10.