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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

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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.

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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.