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Calculate pH of - 0.01 N HCl...

Calculate pH of -
0.01 N HCl

Text Solution

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Strong acids ionize completely at normal dilutions
`0.01 N HCl HCl to H^(+)+Cl^(-)`
Conc. Before ionisation `10^(-2)N" 0 0"`
Conc. After ionisation `"0 "10^(-2)" "10^(-2)`
`[H^(+)]=10^(-2)M`
`pH=-log[H^(+)]`
`? pH =2`
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