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.

Options

A. `11.30`
B. `10.53`
C. `2.70`
D. `8.35`

Answer

A

Explanation

`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`.