You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Nucleophilic Addition at the Carbonyl

What volume o...

Question

What volume of 0.1M KOH is needed to completely neutralise the HCl in 300ml of HCl with Ph =2.25

Open in App

Solution

KOH :
M1 = 0.1 M
V1 = ?

HCl :
pH = 2.25
pH = -log [H+] = -log M2
-log M2 = 2.25
log M2 = -2.25
M2 = 10^(-2.25) = 0.00562
V2 = 300 ml

We know that,
M1V1 = M2V2
0.1×V1 = 0.00562×300
V1 = (0.00562×300)/0.1
V1 = 1.686/0.1
V1 = 16.86 ml.

i.e 16.86 ml of KOH is required.

Suggest Corrections

Similar questions

Q. What will be the resultant

p H

when

200 m L

of an aqueous solution of

H C l

(p H = 2.0)

is mixed with

300 m L

of an aqueous solution of

H C l

(p H = 4.0)

30.0 mL of the given $H C l$ solution requires 20.0 mL of 0.1 M sodium carbonate solution for complete neutralisation. What is the volume of this $H C l$ solution required to neutralise 30.0 mL of 0.2 M $N a O H$ solution?

S O_{2} C l_{2}

(sulphuryl chloride) reacts with water to give a mixture of

H_{2} S O_{4}

and

H C l

. What volume of

0.2 M B a (O H)_{2}

is needed to completely neutralize

25

mL of

0.2 M S O_{2} C l_{2}

solution.

Q. 10mL of a solution of NaOH is found to be completely neutralised by 8mL of a given solution of HCl. HCl solution required to neutralise 50mLof NaOH will be

Q. If

10 m L

1 M

N a O H

is titrated with

1 M

H C l

to a

p H

2

, what volume of

H C l

has to be added?

Join BYJU'S Learning Program

What volume of 0.1M KOH is needed to completely neutralise the HCl in 300ml of HCl with Ph =2.25

KOH : M1 = 0.1 M V1 = ? HCl : pH = 2.25 pH = -log [H+] = -log M2 -log M2 = 2.25 log M2 = -2.25 M2 = 10^(-2.25) = 0.00562 V2 = 300 ml We know that, M1V1 = M2V2 0.1×V1 = 0.00562×300 V1 = (0.00562×300)/0.1 V1 = 1.686/0.1 V1 = 16.86 ml. i.e 16.86 ml of KOH is required.

KOH :
M1 = 0.1 M
V1 = ?

HCl :
pH = 2.25
pH = -log [H+] = -log M2
-log M2 = 2.25
log M2 = -2.25
M2 = 10^(-2.25) = 0.00562
V2 = 300 ml

We know that,
M1V1 = M2V2
0.1×V1 = 0.00562×300
V1 = (0.00562×300)/0.1
V1 = 1.686/0.1
V1 = 16.86 ml.

i.e 16.86 ml of KOH is required.