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# 100 ml of $\text{ 0}\text{.2 M HCl }$ are mixed with 100 ml of $\text{ 0}\text{.2 M C}{{\text{H}}_{\text{3}}}\text{COOH }$, the $\text{ pH }$ of the resulting solution should be nearly:A) $\text{ 1}\text{.0 }$B) $\text{ 0}\text{.7 }$ C) $\text{ 1}\text{.3 }$D) $\text{ 1}\text{.6 }$

Last updated date: 20th Jun 2024
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Hint: Hydrochloric acid and sodium acetate are the strong acids and strong bases respectively. When mixed react with each other to form a salt,
$\text{ C}{{\text{H}}_{\text{3}}}\text{COONa }+\text{ HCl }\to \text{ NaCl + C}{{\text{H}}_{\text{3}}}\text{COOH }$
The concentration of hydrogen ions is equal to the concentration of acetic acid in the total volume of the solution. The $\text{ pH }$ of the solution is written as,
$\text{ pH }=-\log \left[ {{\text{H}}^{\text{+}}} \right]\text{ }$

Complete step by step answer:
To solve this question we will consider that hydrochloric acid reacts with the sodium acetate to form a slat and the acetic acid. the given volume of the acetic acid is generated when the hydrochloric acid reacts with the sodium acetate.
$\text{ C}{{\text{H}}_{\text{3}}}\text{COONa }+\text{ HCl }\to \text{ NaCl + C}{{\text{H}}_{\text{3}}}\text{COOH }$
Therefore,
The volume of the hydrochloric acid solution$\text{ HCl }$ is $\text{ }{{\text{V}}_{\text{HCl}}}\text{ = 100 mL }$
The concentration of hydrochloric acid $\text{ HCl }$is $\text{ }{{\text{C}}_{\text{HCl}}}\text{ = 0}\text{.2 M }$
The volume of the hydrochloric acid solution$\text{ C}{{\text{H}}_{\text{3}}}\text{COONa }$ is $\text{ }{{\text{V}}_{\text{ C}{{\text{H}}_{\text{3}}}\text{COONa }}}\text{ = 100 mL }$
The concentration of hydrochloric acid $\text{ C}{{\text{H}}_{\text{3}}}\text{COONa }$is $\text{ }{{\text{C}}_{\text{ C}{{\text{H}}_{\text{3}}}\text{COONa }}}\text{ = 0}\text{.2 M }$
We are interested to determine $\text{ pH }$ of the solution.
The sodium salt of acetic acid or sodium acetate is a strong base. It undergoes the dissociation in its acetate ion and sodium ion .the reaction of dissociation of sodium acetate is as shown below,
$\text{ C}{{\text{H}}_{\text{3}}}\text{COONa }\to \text{ C}{{\text{H}}_{\text{3}}}\text{CO}{{\text{O}}^{-}}\text{ + N}{{\text{a}}^{\text{+}}}\text{ }$
Here, if we consider that the 100 ml of $\text{0}\text{.2 M }$ sodium acetate is undergoing the dissociation, then the number of moles of sodium ion $\text{ N}{{\text{a}}^{\text{+ }}}$ is equal to the,
$\text{ moles of N}{{\text{a}}^{\text{+}}}\text{ = }{{\text{C}}_{\text{C}{{\text{H}}_{\text{3}}}\text{COONa }}}\text{ }\!\!\times\!\!\text{ }{{\text{V}}_{\text{C}{{\text{H}}_{\text{3}}}\text{COONa}}}=\text{ 0}\text{.2}\times \text{100 = 20 millimolar }$
Thus, the sodium acetate solution will generate 20 mill moles of sodium ion.
Similarly hydrochloric acid $\text{ HCl }$dissociates as follows,
$\text{ HCl }\to \text{ }{{\text{H}}^{\text{+}}}\text{ + C}{{\text{l}}^{-}}\text{ }$
Here, if we consider that the 100 ml of $\text{0}\text{.2 M }$ $\text{ HCl }$is undergoing the dissociation, then the number of moles of chloride ion $\text{ C}{{\text{l}}^{-}}\text{ }$ is equal to the,
$\text{ moles of C}{{\text{l}}^{-}}\text{ = }{{\text{C}}_{\text{HCl }}}\text{ }\!\!\times\!\!\text{ }{{\text{V}}_{\text{HCl}}}=\text{ 0}\text{.2}\times \text{100 = 20 millimolar }$
Thus, the hydrochloric acid solution will generate 20 mill moles of chloride ion in the solution.
These 20 mill moles of sodium ion and 20 mill moles of the chloride react to general the sodium chloride salt in the solution. the reaction between sodium ion and the chloride ion is as shown below,
$\text{ N}{{\text{a}}^{\text{+}}}\text{ + C}{{\text{l}}^{-}}\to \text{ NaCl }$
From the dissociation of hydrochloric acid, we know that the solution contains 20 mill moles of the hydrogen ion $\text{ }{{\text{H}}^{\text{+}}}\text{ }$ in the mixture of solution. Thus the concentration of the hydrogen ion would be equal to the number of moles of hydrogen ion present in the total volume.it is written as follows,
$\text{ }\left[ {{\text{H}}^{\text{+}}} \right]\text{ = }\dfrac{\text{moles}}{\text{ Total volume }}\text{= }\dfrac{\text{20}}{\text{100+100}}\text{= 0}\text{.1 M }$
Since we know that the $\text{ pH }$ negative log value of the hydrogen ion concentration. The $\text{ pH }$of the mixture is calculated as, $\text{ pH }=-\log \left[ {{\text{H}}^{\text{+}}} \right]\text{ = }-\log \left[ 0.1 \right]\text{ = 1 }$
Thus, $\text{ pH }$of the mixture of $\text{ HCl }$and $\text{C}{{\text{H}}_{\text{3}}}\text{COONa}$ is equal to 1.

Hence, (A) is the correct option.

Note: Students should note that sodium acetate and hydrochloric acid are strong electrolytes and it completely dissociates into the solution, however, the acetic acid is a weak electrolyte it dissociates to less extent. On knowing the mill moles we can determine the concentration of the solution. Remember that, here reactions are balanced thus one mole of reactant is given one mole of product. However in the case of other reactions we need to consider the coefficients.