Calculate the pH of the following solutions.
a. 0.37g fo `Ca(OH)_(2)` dissolved in water to give 500 ml solution
b. 0.3 g of NaOH dissolved in water to give 200 ml solution
c. 0.1825% HCl aqueous solution
d. 1 ml of 13.6 M HCl is diluted with water to give 1 litre solution.

Explanation

a. Molarity of `Ca(OH)_(2)`
`=0.37/74xx100/500=0.01`
`Ca(OH)_(2)toCa^(2+)+2OH^(-)`
`0.01, 0.02`
`pOH=-log[OH^(-)]`
`=-log[2xx10^(-2)]=1.6990`
`pH=14-1.6990=12.3010`
b. Molarity of `NaOH=0.3/40xx1000/200=0.0375`
`pOH=-log[3.75xx10^(-2)]=1.426`
`pH=14-1.426=12.574`
c. `M=0.1825/36.5xx1000/100=0.05`
`pH=-log[5xx10^(-2)]=1.3`
d. `M_(1)=13.6,M_(2)=?`
`V_(1)=1, V_(2)=1000`
`M_(2)=(13.6xx1)/1000=1.36xx10^(-2)`
`pH=-log[1.36xx10^(-2)]=1.86`