200 mL of a N10 H2 SO4 solution is mixed with 300 mL of a N100 NaOH solution- Calculate the normality of the excess H-OH- ions in the resulting solution-

Number of gram equivalents of H_2 SO_4 = Normality × Volume in L = 0.1 × nbsp; 0.2 = 0.02 Number of gram equivalents of NaOH = Normality × Volume in L nbsp; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Salt of Strong Acid and Strong Bases

200 mL of a N...

Question

200 mL of a $\frac{N}{10}$ $H_{2} S O_{4}$ solution is mixed with 300 mL of a $\frac{N}{100}$ $N a O H$ solution. Calculate the normality of the excess $H^{+} / O H^{-}$ ions in the resulting solution.

0.024 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.015 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.050 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.034 N

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $0.034 N$
Number of gram equivalents of $H_{2} S O_{4} =$ Normality $\times$ Volume in L
$= 0.1 \times 0.2 = 0.02$
Number of gram equivalents of NaOH = Normality $\times$ Volume in L
$= 0.01 \times 0.3 = 0.003$
0.003 gram equivalents of $N a O H$ will neutralize 0.003 gram equivalents of $H_{2} S O_{4} .$
Hence, the number of gram equivalents of $H_{2} S O_{4}$ in 500 mL of the solution after neutralisation = 0.02 - 0.003
= 0.017
Hence, the number of gram equivalents of $H^{+}$ = 0.017
$Normality of H^{+} = \frac{Number of gram equivalents}{Volume of solution in L}$

Normality of $H^{+}$ = $\frac{0.017}{0.5} = 0.034 N$

Suggest Corrections

Similar questions

Q. 200 mL of a

\frac{N}{10}

H_{2} S O_{4}

solution is mixed with 300 mL of a

\frac{N}{100}

NaOH solution. Calculate the normality of the excess

H^{+} / O H^{-}

ions in the resulting solution.

200 mL

0.1 M

H_{2} S O_{4}

is added to

300 mL

0.1 M

N a O H

solution. Find the normality of the ions in excess (

H^{+}

O H^{-}

) in the resultant solution.

Q. The normality of the excess ions

(H^{+} / O H^{-})

in a solution obtained by mixing

100 mL

of a

0.2 M H_{2} S O_{4}

solution with

100 mL

of a

0.2 M N a O H

solution is:

Q. 20 mL of a 10 N HCl solution is mixed with 10 mL of a 36 N

H_{2} S O_{4}

solution and this solution is further diluted with water to 1 L. The normality of

H^{+}

ions in the resultant solution will be:

Q. Calculate molarity of the resulting mixture, when 0.1 N of 100 ml

H_{2} S O_{4}

is mixed with 0.2 N of 100 ml of NaOH solution:

Join BYJU'S Learning Program

200 mL of a N10 H2SO4 solution is mixed with 300 mL of a N100 NaOH solution. Calculate the normality of the excess H+/OH− ions in the resulting solution.

200 mL of a $\frac{N}{10}$ $H_{2} S O_{4}$ solution is mixed with 300 mL of a $\frac{N}{100}$ $N a O H$ solution. Calculate the normality of the excess $H^{+} / O H^{-}$ ions in the resulting solution.