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Question: Equal volumes of two HCl solutions of \[pH=3,pH=5\] were mixed. What is the pH of the resulting solu...

Equal volumes of two HCl solutions of pH=3,pH=5pH=3,pH=5 were mixed. What is the pH of the resulting solution?

Explanation

Solution

We know that pH is the criteria which measure whether the given compound, solution, etc. falls under the category of acid or base. Acids are those which are capable of losing hydrogen ions and bases are those which are capable of losing ion in the solution.

Complete step by step solution:
pH of solution is defined as the decimal logarithm of the reciprocal of hydrogen ion activity, aH+{{a}_{{{H}^{+}}}}
pH=−log⁡10(aH+)=log⁡10(1aH+)pH=-{{\log }_{10}}\left( {{a}_{{{H}^{+}}}} \right)={{\log }_{10}}\left( \dfrac{1}{{{a}_{{{H}^{+}}}}} \right)
pH of solution changes with temperature, that is it depends on the temperature for example pH of water at 00 degree Celsius is , 7.477.47 pH of water at 25∘C25{}^\circ C is 77 and at 100∘C100{}^\circ C is 6.146.14 . This is because as the temperature increases most of the water molecules converted to vapour and concentration of the water increases due to which pH changes. This definition of pH was adopted because ion-selective electrodes, which are used to measure pH, respond to activity;
pH scale ranging number from 0 to 14.0\text{ }to\text{ }14.
If pH<7pH<7 then it is said to be an acid.
If pH>7pH>7 then it is said to be base.
If pH=7pH=7 then it is neutral.
This is because as the temperature increases most of the water molecules converted to vapour and concentration of the water increases due to which pH changes.
pH=−log⁡(H+)=10−pHpH=-\log ({{H}^{+}})={{10}^{-pH}}
Moles of solution I=10−3×V=10−3VI={{10}^{-3}}\times V={{10}^{-3}}V
Moles of solution II=10−5×V=10−5VII={{10}^{-5}}\times V={{10}^{-5}}V
Thus, [H+]=10−3V+10−3V2V[{{H}^{+}}]=\dfrac{{{10}^{-3}}V+{{10}^{-3}}V}{2V}
⇒[H+]=100×10−5+10−52\Rightarrow [{{H}^{+}}]=\dfrac{100\times {{10}^{-5}}+{{10}^{-5}}}{2}
⇒[H+]=101×10−52=50.5×10−5\Rightarrow [{{H}^{+}}]=\dfrac{101\times {{10}^{-5}}}{2}=50.5\times {{10}^{-5}}
Thus, pH=−log⁡(5.05×10−5)=4−log⁡5.05pH=-\log (5.05\times {{10}^{-5}})=4-\log 5.05
pH=4−0.7=3.3pH=4-0.7=3.3
Therefore, the pH of resulting solution is 3.33.3

Note: To solve this kind of problem we must know the pH scale and the values which fall under the acid, Base and neutral. The pH of water is seven then it is neutral in nature. The pH scale tells us about the concentration of hydrogen ions present in the aqueous solution of the given compound.