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

Byju's Answer

Standard XII

Chemistry

Dissolution of Solids and Gases in Liquid

If 10 mL of...

Question

If $10 m L$ of $1 M$ $N a O H$ is titrated with $1 M$ $H C l$ to a $p H$ of $2$ , what volume of $H C l$ has to be added?

10.2 m L

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

13.2 m L

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

18.2 m L

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

20.2 m L

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

Open in App

Solution

The correct option is A $10.2 m L$
$N_{1} V_{1}     $ $=$ $N_{2} V_{2}     $ $∵ [H^{+}] = 10^{- 2} M$ (Given)
$N a O H$ $H C l$ Initial $[H^{+}] = 1 M$
So, ${c o n c}^{n}$ $H^{+}$ consumed $= (1 - 10^{- 2})$
$10 \times 1 = (1 - 10^{- 2}) . V_{2}$
$V_{2} = (\frac{10}{1 - 0.01}) = 10.10 / m l \approx 10.2 m l$

Suggest Corrections

Similar questions

Q. A solution of weak acid

H A

was titrated with base

N a O H

. The equivalence point was reached when

36.12 m L

0.1 M

N a O H

has been added. Now

18.06 m L

0.1 M

H C l

were added to titrated solution, the

p H

was found to be

4.92

. What will be the

p H

of the solution obtained by mixing

10 m L

0.2 M

N a O H

and

10 m L

0.2 M

H A

Q. A

30 m L

solution of

0.2 M N H_{4} O H

is titrated with

0.2 M H C l

. Find the

p H

of the solution when

10 m L

H C l

has been added.
Given

p K_{b} (N H_{4} O H) = 4.74

Q. A student sets up a titration of

40

mL of

1 M

N a O H

against

2 M

H C l

, determine the volume of

H C l

required to neutralize?

Q. What is the concentration if

10 m L

H C l

which was titrated with

40 m L

0.10 M

N a O H

Q. In an experiment,

20 m L

of decinormal

H C l

solution was added to

10 m L

of a decinormal

A g N O_{3}

solution.

A g C l

was precipitated out and the excess acid was titrated against a decinormal

N a O H

solution. Volume of

N a O H

required for this back titration is:

Join BYJU'S Learning Program

If 10mL of 1M NaOH is titrated with 1M HCl to a pH of 2, what volume of HCl has to be added?

The correct option is A 10.2mL N1V1 = N2V2 ∵[H+]=10−2M (Given)NaOH HCl Initial [H+]=1M So, concn H+ consumed =(1−10−2)10×1=(1−10−2).V2V2=(101−0.01)=10.10/ml≈10.2ml

If $10 m L$ of $1 M$ $N a O H$ is titrated with $1 M$ $H C l$ to a $p H$ of $2$ , what volume of $H C l$ has to be added?