1.

Determine pH of the solution that results from addition of 20 ml of 0.01 M Ca(OH)_(2) to 30 ml of 0.01 M HCl.

Answer»

`11.30`
`10.53`
`2.70`
`8.35`

Solution :`Ca(OH)_(2)+2HCl rarr CaCl_(2)+2H_(2)O`
Moles of `Ca(OH)_(2)` in 20 ml solution
`=(0.01)/(1000)xx20=2xx10^(-4)` mole
Moles of HCl in 30 ml solution
`=(0.01)/(1000)xx30`
`=3.0xx10^(-4)`mole.
2 moles of HCl react with 1 mole of `Ca.(OH)_(2)`.
Hence `3xx10^(-4)`moles of HCl react with 1.5 moles of `Ca(OH)_(2)`
Moles of `Ca(OH)_(2)` left
`=2xx10^(-4)-1.5xx10^(-4)=0.5xx10^(-4)`
`=5xx10^(-5)` moles
Moles of `OH^(-)=2xx5xx10^(-5)=10^(-4)`moles.
`[OH^(-)]=(10^(-4))/(50)xx1000=2xx10^(-3)M`
`pOH=-LOG[OH^(-)]=-log(2xx10^(-3))`
`=2.699`
`pH=14-pOH=11.301`.


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