10 mL of a N -2 HCl solution- 30 mL of a N -10 HN O 3 solution and 75 mL of a N -5 HNO 3solution are mixed- The normality of H -ions in the resultant solution will be-A- 0-1 NB- 0-4 NC- 0-2 ND- 0-5 N

The correct option is C 0.2 NEquivalents of H^+ in 10 mL of (N/2) nbsp; HCl=10/2× 10^ 3 Equivalents of H^+ in 30 mL of (N/10) nbsp; HNO_3= 30/10× 10^ 3 Equ ...

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Byju's Answer

Standard XII

Chemistry

Normality

10 mL of a N ...

Question

10 mL of a $\frac{N}{2}$ HCl solution, 30 mL of a $\frac{N}{10} H N O_{3}$ solution and 75 mL of a $\frac{N}{5} H N O_{3}$
solution are mixed. The normality of $H^{+}$ ions in the resultant solution will be:

0.2 N

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0.1 N

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0.5 N

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0.4 N

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Solution

The correct option is A 0.2 N
Equivalents of $H^{+}$ in 10 mL of $(\frac{N}{2})$ HCl $= \frac{10}{2} \times 10^{- 3}$
Equivalents of $H^{+}$ in 30 mL of $(\frac{N}{10}) H N O_{3} = \frac{30}{10} \times 10^{- 3}$
Equivalents of $H^{+}$ in 75 mL of $(\frac{N}{5}) H N O_{3} = \frac{75}{5} \times 10^{- 3}$
Hence, total equivalents of $H^{+} = (5 + 3 + 15) \times 10^{- 3} = 23 \times 10^{- 3}$
Total volume of solution = 115 mL
Hence, normality of $H^{+}$ in the resultant solution $= \frac{23 \times 10^{- 3} \times 1000}{115} = \frac{N}{5} = 0.2$ N

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10 mL of a N2 HCl solution, 30 mL of a N10HNO3 solution and 75 mL of a N5HNO3 solution are mixed. The normality of H+ ions in the resultant solution will be:

10 mL of a $\frac{N}{2}$ HCl solution, 30 mL of a $\frac{N}{10} H N O_{3}$ solution and 75 mL of a $\frac{N}{5} H N O_{3}$
solution are mixed. The normality of $H^{+}$ ions in the resultant solution will be: