The ph of strong acid is 5 and it is diluted 100 times.what will be the ph of obtained solution?
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according to Arhenius rule
Ph = - log [H+]
where [H+] is concentration of acid
initially ,
5 = - log [H+]
[H+] = 10^-5
hence initial concentration of strong acid is 10^-5 .
now ,
when we dilute 100 times then concentration decreases 100 times .
because concentration inversely proportional to volume .
now,
final [H+] =10^-5/100 = 10^-7
now,
Ph =-log [H+]
= -log (10^-7 ) =7
hence ,
Ph =7
Ph = - log [H+]
where [H+] is concentration of acid
initially ,
5 = - log [H+]
[H+] = 10^-5
hence initial concentration of strong acid is 10^-5 .
now ,
when we dilute 100 times then concentration decreases 100 times .
because concentration inversely proportional to volume .
now,
final [H+] =10^-5/100 = 10^-7
now,
Ph =-log [H+]
= -log (10^-7 ) =7
hence ,
Ph =7
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