You visited us 6 times! Enjoying our articles? Unlock Full Access!

Atomic Valency

Among the fol...

Question

Among the following, the plot that correctly represents the conductometric titration of $0.05 M H_{2} S O_{4}$ with $0.1 M N H_{4} O H$ is,

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

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

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

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

Open in App

Solution

The correct option is D
Initially $H^{+}$ ions are present due to complete dissociation of $H_{2} S O_{4}$ . Upon addition of $N H_{4} O H$ , $H^{+}$ ions decreases, so conductivity decreases and becomes constant after salt formation.

Suggest Corrections

Similar questions

0.1 {M Ba(OH)}_{2}

0.1 {M MgSO}_{4}

{MgSO}_{4}

H C l and N a O H

100 mL

0.1 M NaOH

100 mL

0.05 M H_{2} S O_{4}

p H

(for H_{2} S O_{4}, K a_{1} = \infty, K a_{2} = 10^{- 2})

20 m L

0.1 M

H_{2} S O_{4}

30 m L

0.2 M

N H_{4} O H

p H

p K_{b}

N H_{4} O H = 4.7

Join BYJU'S Learning Program