500 ml of 0.1 M KCl, 200 ml of 0.01 M $N a N O_{3}$ and 500 ml of 0.1 M $A g N O_{3}$ was mixed. What is the total number of moles of ions in solution?

Q. 500 ml of a 0.1 N solution of

A g {N O}_{3}

is added to 500 ml of 0.1 N solution of KCl. The concentration of nitrate ion in the resulting mixture is
(a) 0.05 N (b) 0.1 N (c) 0.2 N (d) reduced to zero

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500 mL of 0.1 M H2SO4 is added to 500 mL of 0.1 M NaOH solution. Find the normality of the ions in excess (H+ or OH−) in the resultant solution.

The correct option is D 0.05 NMilliequivalents of NaOH =500×0.1×1=50 Milliequivalents of H2SO4=500×2×0.1=100 Since, the acid is present in excess, the solution is acidic. Hence, milliequivalents of excess H+ = 100 – 50 = 50 Normality of H+=501000=0.05 N

500 mL of 0.1 M $H_{2} S O_{4}$ is added to 500 mL of 0.1 M $N a O H$ solution. Find the normality of the ions in excess ( $H^{+}$ or $O H^{-}$ ) in the resultant solution.