Question

Calculate the \[\left[ {C{l^ - }} \right]\] , \[\left[ {N{a^ + }} \right]\] , \[\left[ {{H^ + }} \right]\] , \[\left[ {O{H^ - }} \right]\] and pH of the resulting solution obtained by mixing \[50mL\] of \[0.6N\] \[HCl\] and \[50mL\] of \[0.3N\] \[NaOH\] .

Answer 1

Hint: We need to know the definition of pH and the concept of normality and molarity. Molarity, M is defined as the number of moles of solute per liter of solution and is given by the formula:
$M = \dfrac{{{\text{Number of moles of solute}}}}{{{\text{volume of solution in litres}}}}$ .
Normality, N is described as the number of gram or mole equivalents of solute present in one litre of a solution. For a monovalent compound, i.e, compounds whose constituent elements have a valency of 1, Normality=Molarity. Also, $pH = - \log \left[ {{H^ + }} \right]$ .

Complete step by step answer:
From the given question:
Volume of \[HCl = 50ml\]
Concentration of \[HCl = 0.6N\]
Therefore, number of moles of \[HCl\] = Normality X volume of solution = \[0.6 \times 50 = 30\]
Similarly,
Volume of \[NaOH = 50mL\]
Concentration of \[NaOH = 0.3N\]
Therefore, number of moles of \[NaOH\] = Normality X volume of solution =\[0.3 \times 50 = 15\]
When \[50mL\] of \[0.6N\]\[HCl\] and \[50mL\] of \[0.3N\] \[NaOH\] are mixed, the following reaction takes place: \[HCl + NaOH \to NaCl + {H_2}O\]
No. Of moles of \[HCl\] involved in the initial reaction when no products are formed = 30
No. Of moles of \[NaOH\] involved in the initial reaction when no products are formed = 15
 Since 15 moles of \[NaOH\] are involved, it is capable of consuming only 15 moles.
Therefore, in the reactants:
No. Of moles of \[HCl\] left after the products are formed=15
No. Of moles of \[NaOH\] left after the products are formed =0
In the products:
No. Of moles of \[NaCl\] formed = 15
No. Of moles of \[{H_2}O\] formed =15
Total volume after the reaction \[ = 50mL + 50mL\]
Total volume after the reaction is $100mL$ .
Hence,
Concentration of \[C{l^ - }\], $\left[ {C{l^ - }} \right] = \dfrac{{{\text{Number of moles of solute}}}}{{{\text{volume of solution in litres}}}}$ = $\dfrac{{15 + 15}}{{100}}$ ( \[15\] moles from \[HCl\] and \[15\] moles from \[NaCl\])
Concentration of $C{l^ - }$ = $\dfrac{{30}}{{100}} = 0.3M$
Concentration of \[N{a^ + }\] , $\left[ {N{a^ + }} \right] = \dfrac{{{\text{Number of moles of solute}}}}{{{\text{volume of solution in litres}}}}$ = $\dfrac{{15}}{{100}}$ ( \[15\] moles from \[NaCl\] )
Concentration of $N{a^ + } = 0.15M$
Concentration of \[{H^ + }\], $\left[ {{H^ + }} \right] = \dfrac{{{\text{Number of moles of solute}}}}{{{\text{volume of solution in litres}}}}$=$\dfrac{{15}}{{100}}$( \[15\] moles from \[NaCl\] )
Concentration of ${H^ + } = 0.15M$
Concentration of \[O{H^ - }\], $\left[ {O{H^ - }} \right] = \dfrac{{{\text{Number of moles of solute}}}}{{{\text{volume of solution in litres}}}}$= \[\dfrac{{{{10}^{ - 14}}}}{{0.15}}\] (in terms of water)
Concentration of $O{H^ - } = 6.6 \times {10^{ - 14}}M$
Also, pH of the resulting reaction=\[pH = - log\left[ {{H^ + }} \right] = - log0.15 = 0.8239\] .

Note: We must be noted that Molarity is a measurement of the moles in the total volume of the solution, whereas Normality is a measurement of the gram equivalent in relation to the total volume of the solution. The number of moles is also called milliequivalent. Also, the concentration of the constituent atoms can be expressed both in terms of molarity and normality as it is a monovalent compound.