Question

Calculate the pH of the mixture :$8$ g of $NaOH$ $+ 680$ mL of $1$ M $HCl$+$10$ mL of ${H_2}S{O_4}$ ​, (specific gravity $1.2$, $49\%$ of ${H_2}S{O_4}$​ by mass). The total volume of the solution was made to $1$ Litre with water:
A.$1 + 2\log 3$
B.$1 - \log 6$
C.$3 + 3\log 3$
D.$3 - 2\log 6$

Answer 1

The concentration of a solution is commonly expressed by its molarity, defined as the amount of dissolved substance per unit volume of solution, for which the unit used in moles per litre. The unit of molarity is $mol/L$. A$1$ M solution is said to be “one molar.”
Formula used:
$n = \dfrac{m}{M}$
where,
$n =$ number of moles
$m =$ mass of a substance
$M =$ molar mass
$Molarity = \dfrac{n}{L}$
where,
$n =$ number of moles of solute
$L =$ volume of solution in litres
To calculate the pH of an aqueous solution, you need to know the concentration of hydronium ions in moles per litre (molarity) of the solution.pH is then calculated using this expression
$pH = - {\log _{10}}\left[ {{H_3}{O^ + }} \right]$

Complete answer:
Given that
Mass of $NaOH = 8g$
Volume of $HCl$ solution $= 680ml$
To convert ml to litre, we divide by $1000$
$\dfrac{{680}}{{1000}} = 0.68L$
Molarity of $HCl$ solution, $= 1M$
Volume of ${H_2}S{O_4}$ solution ​$= 10ml$
Density of ${H_2}S{O_4}$ solution ,$= 1.2\,g/ml$
Mass $\%$ of ${H_2}S{O_4}$​ in solution $= 49\%$
Now,
(1) $NaOH \to N{a^ + } + O{H^ - }$
Moles of $O{H^ - }$ $=$ Moles of $NaOH$
We know, molar mass of $NaOH$ is $40$g
So ,on substituting the values in the formula of number of moles $n = \dfrac{m}{M}$ , we get
$n = \dfrac{8}{{40}}$
$\Rightarrow n = 0.2$
(2) $HCl \to {H^ + } + C{l^ - }$
We know, $Molarity = \dfrac{n}{L}$
Moles of ${H^ + }$ from $=$ Moles of $HCl$ $=$Molarity $\times$ Volume
=$1M \times 0.68lit = 0.68mol$
(3) Mass of ${H_2}S{O_4}$ (solution)
Since specific gravity is mentioned, mass will be the product of specific gravity and volume of sulphuric acid solution, that is
$\Rightarrow 1.2 \times 10$
$m = 12g$
Mass of ${H_2}S{O_4}$

$$= 49\% \;of\;12g \\\ = 0.49 \times 12 \\\ = 5.88g \\\$$

3. ${H_2}S{O_4} \to 2{H^ + } + SO_4^{2 - }$​
   Moles of hydrogen ions from ${H_2}S{O_4}$ $= 2 \times$ Moles of ${H_2}S{O_4}$​
   Molar mass of ${H_2}S{O_4}$=$98g$
   So, no of moles$= 2 \times \dfrac{{5.88}}{{98}}$
   $n = 0.12mol$
   Therefore, Total moles of ${H^ + }$ in solution is given by
   $0.68 + 0.12 = 0.8mol$
   Now,
   Acid base reaction
   {H^ + } + O{H^\\_} \to H2O
   Moles of excess / unreacted hydrogen ions
   The leftover moles which are in excess are given by
   $0.8 - 0.2 = 0.6mol$
   Molarity of Hydrogen ions in the solution will be equal to the number of moles of unreacted hydrogen ions divided by the volume of final solution
   Molarity of $\left[ {H + } \right] = \dfrac{{0.6}}{1} = 0.6M$
   Now,
   $Ph = - log\left[ {H + } \right] = - log\left( {0.6} \right) = - log\left( {106} \right) = - log\left( 6 \right) - log\left( {10} \right)$
   $= - log6 + log10 = - log 6 + 1 = 1 - log6$
   The correct answer is option (B)$1 - log6$

Note:
-The molar mass of a compound is equal to the sum of the atomic masses of its constituent atoms in g/mol.
-Although there is no physical way of measuring the number of moles of a compound, we can relate its mass to the number of moles by using the compound’s molar mass as a direct conversion factor.
-To convert between mass and number of moles, you can use the molar mass of the substance.