Question

The normality of a solution that results from mixing $4$ g of NaOH, $500$ mL of $1$ M HCl, and $10.0$ mL of $H_{2}S{O}_4$ (specific gravity = $1.1$, $49$% $H_{2}S{O}_4$ by weight) is :
(The total volume of solution was made to $1$ L with water)

• A. 0.51
• B. 0.71
• C. 1.02
• D. 0.45

Answer 1

The correct option is A 0.51

Mol. wt of NaOH $=23+16+1=40g/mol$
Moles of NaOH =Equivalents of NaOH$=4/40=0.1moles$
Moles of HCl=Equivalents of HCl
$=\frac{Vol.\left(mL\right)\times Molarity}{1000}=\frac{500\times 1}{1000}=0.5$
Mass of $H_{2}S{O}_4$ and water in soln. $=10\times 1.1=11g$
Mass of $H_{2}S{O}_4$in soln.
$=0.49\times 11=5.39g$
Mol. wt of $H_{2}S{O}_4=2+32+64=98g/mol$
Moles of $H_{2}S{O}_4=5.39/98=0.055moles$
Equivalents of $H_{2}S{O}_4=0.055\times 2=0.11$
Total no. of acid equivalents $=0.5+0.11=0.61$
Total no. of base equivalents $=0.1$
Base (NaOH) is limiting reactant and it will react completely upon mixing.
Remaining equivalents of acid after mixing $=0.61-0.1=0.51$
Total equivalents after mixing $=0.51$
$Normality=\frac{No.ofequivalents}{Vol.ofsoln.\left(L\right)}$
$=\frac{0.51}{1}=0.51N$
Thus (a) is the correct answer