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% $H_{2}S{O}_4$ by mass). The total volume of the solution was made to 1 L with water:

• A. 1+2 log 3
• B. 1-log 6
• C. 3 + 3 log 3
• D. 3-2 log 6

Answer 1

The correct option is B 1-log 6

Given that

Mass of $NaOH,{}^{W}NaOH=8g$

Volume of $HClsoln,{}^{V}HCl=680ml=0.68ml$

Molaqrity of $HCl$ soln, ${}^{M}HCl=1M$

Volume of $H_{2}S{O}_{4}soln.{}^V{H}_{2}S{O}_4=10ml$

Density of $H_{2}S{O}_{4}soln{,}^S{H}_{2}S{O}_4=1.2g/ml$

Mass $\%$ of $H_{2}S{O}_4$ in soln=$49\%$

Now,

(1) $NaOH\to N{a}^{+}+O{H}^{-}$

Moles of $O{H}^{-}$=Moles of $NaOH=\frac{{}^{W}NaOH}{{}^{M}NaOH}=\frac{8g}{40g/mol}=0.2moles$

(2) $HCl\to H^{+}+C{l}^{-}$

Moles of $H^{+}$ from =Moles of HCl=Molarity $\times$ Volume

$=1M\times 0.68lit=0.68mol$

(3) Mass of $H_{2}S{O}_4\left(soln\right){=}^S{H}_{2}S{O}_4{\times }^V{H}_{2}S{O}_4=1.2g/mol\times 10mole$

$=12g$

Mass of $H_{2}S{O}_4=49\%$ of $12g=\frac{49}{100}\times 12g=5.88g$

$H_{2}S{O}_4\to 2H^{+}+S{O}_{+}^{2-}$

Moles of $H^{+}$ form $H_{2}S{O}_4=2\times Molesof{H}_{2}S{O}_4=2\times \frac{{}^W{H}_{2}S{O}_4}{{}^M{H}_{2}S{O}_4}$

$=2\times \frac{5.88g}{98g/mol}=0.12mol$

$\therefore$ Total moles of $H^{+}$ in soln =${}^n{H}^{+}(HCl{+}^{n}H+(H_{2}S{O}_4)$

$=(0.68+0.12)mol=0.8mol$

Now,

Acid base reaction

$H^{+}+O{H}^{-}\to H_{2}O$

$(0.8-0.2)(0.2-0.2)$

Moles of excess or unreacted $H^{+}=(0.8-0.2)mol=0.6mol$

Molarity of $H^{+}\left[H^{+}\right]=\frac{{}^{n}H+}{{}^{V}mix}=\frac{0.6mol}{1lid}=0.6M$

Now,

$Ph=-log\left[H^{+}\right]=-log\left(0.6\right)=-log\left(\frac{6}{10}\right)=-log\left(6\right)-\log \left(10\right)$

$=-log6+\log 10=-\log 6+1=1-\log 6$

$\left(B\right)1-\log 6$