Question

A 50 ml solution of pH =1 is mixed with a 50 ml solution of pH =2. The pH of the mixture will be nearly:
A. 0.76
B. 1.26
C. 1.76
D. 2.26

Answer 1

Hint: Calculate the concentration of hydrogen ions for both the ${\text{pH}}$. Using the concentration of hydrogen ions and the volume of each solution calculate the moles of hydrogen in each solution. Using the total moles of hydrogen ions and the total volume of the solution calculate the concentration of hydrogen ions in the solution. Finally, calculate the $pH$ of the solution.
We can use the Formula:
$pH = - \log [{{\text{H}}^{\text{ + }}}]$
${\text{Molarity = }}\dfrac{{{\text{moles}}}}{{{\text{L of solution}}}}$

Step by step answer: Calculate the concentration of hydrogen ions for ${\text{pH}}$ 1 as follows:
${\text{pH}} = - \log [{{\text{H}}^{\text{ + }}}]$
$1 = - \log [{{\text{H}}^{\text{ + }}}]$
$[{{\text{H}}^{\text{ + }}}] = {\text{ antilog ( - 1)}}$
$[{{\text{H}}^{\text{ + }}}] = {\text{ 0}}{\text{.1M}}$
Now, calculate the moles of hydrogen ion in 50 ml solution of pH =1 as follows:
${\text{Molarity = }}\dfrac{{{\text{moles}}}}{{{\text{L of solution}}}}$
Volume of solution = 50 ml = 0.05L
${\text{0}}{\text{.1 M = }}\dfrac{{{\text{moles [}}{{\text{H}}^{\text{ + }}}{\text{] ions}}}}{{{\text{0}}{\text{.05 L}}}}$
${\text{moles [}}{{\text{H}}^{\text{ + }}}{\text{] ions = 0}}{\text{.005 mol}}$

Similarly, calculate the concentration of hydrogen ions for ${\text{pH}}$ 2 as follows:
${\text{pH}} = - \log [{{\text{H}}^{\text{ + }}}]$
$2 = - \log [{{\text{H}}^{\text{ + }}}]$
$[{{\text{H}}^{\text{ + }}}] = {\text{ antilog ( - 2)}}$
$[{{\text{H}}^{\text{ + }}}] = {\text{ 0}}{\text{.01M}}$
Now, calculate the moles of hydrogen ion in 50 ml solution of pH =2 as follows:
${\text{Molarity = }}\dfrac{{{\text{moles}}}}{{{\text{L of solution}}}}$
Volume of solution = 50 ml = 0.05L
${\text{0}}{\text{.01 M = }}\dfrac{{{\text{moles [}}{{\text{H}}^{\text{ + }}}{\text{] ions}}}}{{{\text{0}}{\text{.05 L}}}}$
${\text{moles [}}{{\text{H}}^{\text{ + }}}{\text{] ions = 0}}{\text{.0005 mol}}$
Calculate the concentration of hydrogen ions in the mixture as follows.
${\text{ Total moles [}}{{\text{H}}^{\text{ + }}}{\text{] ions = 0}}{\text{.005 mol + 0}}{\text{.0005 mol = 0}}{\text{.0055 mol}}$
Total volume of mixture = 0.05L +0.05L = 0.1L
\[{\text{[}}{{\text{H}}^{\text{ + }}}{\text{] = }}\dfrac{{{\text{0}}{\text{.0055 mol}}}}{{{\text{0}}{\text{.1L}}}}\]
\[{\text{[}}{{\text{H}}^{\text{ + }}}{\text{] = 0}}{\text{.055 M}}\]
Now, we have a concentration of hydrogen ion in the mixture so calculate the pH of the mixture as follows:
${\text{pH}} = - \log [{{\text{H}}^{\text{ + }}}]$
Substitute 0.055 M for \[{\text{[}}{{\text{H}}^{\text{ + }}}{\text{] }}\]and solve for pH of the mixture.
${\text{pH}} = - \log (0.055)$
${\text{pH}} = 1.26$
So, the pH of the mixture will be nearly 1.26.

Thus, the correct option is (B) 1.26.

Note: When two solutions of different pH are mixed together pH of the mixture will be greater than the pH of the solution having a lower pH value and smaller than the pH of the solution having greater pH value. The acidic solution has a pH less than 7 while the basic solution has a pH greater than 8. The pH of a neutral solution is 7.