Calculate the pH of the resultant mixture:
a. `10 mL` of `0.2M Ca(OH)_(2)+25 mL` of `0.1 M HCl`
b. `10 mL` of `0.01 M H_(2)SO_(4) + 10 mL` of `0.01 M Ca(OH)_(2)`.
c. `10 mL` of `0.1 M H_(2)SO_(4)+ 10 mL` of `0.1 M KOH`.

Explanation

(a) Meq. of `Ca(OH)_(2)=10xx0.2xx2=4`
`[Meq. =NxxV_(mL) = Mxx"Valance factor" xx V_(mL]`
`:. "Meq. of HCI" = 25xx0.1xx1=2.5`
`"Meq. of" Ca(OH)_(2) "left"= 4-2.5=1.5`
`:. N_(Ca(OH)_(2))=(1.5)/(10+25)=4.29xx10^(-2)`
`:. [OH^(-)]=4.29xx10^(-2)`
`:. pOH=1.3675`
`:. pH= 12.6325`
(b) `Meq.of H_(2)SO_(4)=10xx0.01xx2=0.2`
`Meq.of Ca(OH)_(2)=10xx0.01xx2=0.2`
Solution is neutral and `pH=7`
(c ) `Meqof H_(a)SO_(4)=10xx0.1xx2=2`
`Meq.of KOH=10xx0.1xx1=1`
`:. Meq.of H_(2)SO_(4)"left"=1`
`:. N_(H_(2)SO_(4))=1//20=5xx10^(-2)`
`:. [H^(+)]=5xx10^(-2)and pH=1.3010`