1.

Which would be decrease the pH of 25 cm^(3) of a 0.01 M solution of hydrochloric acid

Answer»

The ADDITION of `25cm^(3)` 0.005 M hydrochloric acid
The addition of `25cm^(3)` of 0.02 M hydrochloric acid
The addition of MAGNESIUM metal
None of these

Solution :Given concentration of HCL = 0.01 M
`pH = -log 10^(-2) = 2`
In option (b)
Concentration of HCl = 0.02 M
M.eq `= 0.02 xx 25 = .5 M`. Eq of HCl
M.eq in given condition `= 0.01 xx 25 = 0.25`
Total M.eq in solution `= 0.25 + 0.50 = 0.75 = (.75)/(25+25) = 1.5 xx 10^(-2)N`
Then new pH will be
`= -log 1.5 xx 10^(-2) = 2 - log 1.5`
Which is less than 2 ltbr. So we canvery that that by adding `25cm^(3)` of 0.2 M HCl the pH of `25 cm^(3)` of 0.01 M decrease.


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