A sample of water has a hardness expressed as 80 ppm of `Ca^(2+)` This sample is passed through an ion exchange column and the `Ca^(2+)` is replaced by `H^(+)`.What is the pH of the water after it has been so treated ? [Atomic mass of Ca=40]
A. 3
B. 2.7
C. 5.4
D. 2.4

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Correct Answer - D
`10^6` ml water contains 80 gm of `Ca^(2+)=80/40` moles =2 moles of `Ca^(2+)=2xx2` moles of `H^+` ions
so `10^3` ml of `H_2O` will have =`4xx10^(-3)` moles of `H^+` ions
so pH=3-log 4=3-0.6=2.4

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