Question

50 mL of 1 M HCl, 100 mL of 0.5 M $HN{O}_3,$ and x mL of 5 M $H_{2}S{O}_4$ are mixed together and the total volume is made upto 1.0 L with water. 100 mL of this solution exactly neutralises 10 mL of M/3 $A{l}_2(C{O}_3{)}_3.$ Calculate the value of x:___

Answer 1

Total mEq of acid = $50\times 1+100\times 0.5+x\times 5\times 2$ (n factor)
= (100 + 10x)mEq
$=\frac{(100+10x)}{100mL}N$
$N_1{V}_1\left(Acid\right)=N_1{V}_1\left[A{l}_2\right(C{O}_3^{2-}{)}_3]$ (Total charge = 6)
(n = 6)
$\therefore \frac{(100+10x)}{100mL}N\times 100mL=10mL\times \frac{1}{3}\times 6$
(100 + 10x) = 200
$\therefore$ x = 10 mL