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  4. The normality of a solution that results from mixing 4 g of NaOH , 500 mL of 1 M HCl, and 10.0 mL 2 of H 2 O 4 (specific gravity 1.1,49 % H 2 SO 4 by weight) is

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Q. The normality of a solution that results from mixing $4 \,g$ of $NaOH , 500 \,mL$ of $1 \,M\, HCl$, and $10.0\, mL ^2$ of $H _2 O _4$ (specific gravity $1.1,49 \% H _2 SO _4$ by weight) is

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Solution:

$4 \,g$ of $NaOH =4 / 40=0.1 \,mole =100\, mmol \equiv$
$100 \,mEq$
$HCl =500 \times 1=500 \,mEq$
$N$ of $H _2 SO _4=\frac{W_2 \times 1000}{E w_2 \times V_{\text {sol }}( in mL )}$
or $=\frac{\% \text { by weight } \times 10 \times d}{E w_2}$
$=\frac{49 \times 10 \times 1.1}{49}=11 \,N$
mEq of $H _2 SO _4=11 \,N \times 10.00 \,mL =110 \,mEq$
Total acid $=100+500=610\, mEq$
$NaOH =100 \,mEq$
Acid left $=610-100=510\, mEq$
Total volume $=1000\, mL$
Normality of solution $=\frac{ mEq }{ mL }=\frac{510}{1000}=0.51\, N$