Calculate the `pH` of a solution which contains `100mL` of `0.1MHCl` and `9.9mL` of `1.0MNaOH`.
Take `(1)/(1099)=9.1xx10xx10^(-4)`& `log9.1=0.96`.

Explanation

Given is the case of a mixture of a strong acid and a strong base.
`{:(,HCl+,NaOHrarr,NaCl+,H_(2)O),("Millimoles before reaction",100xx0.1,9.9xx1,0,0),(,10,9.9,9.9,9.9),("Millimoles after reaction",0.1,0,9.9,9.9):}`
`:.[H^(+)]`=form left `HCl =(0.1)/(109.9)=9.1xx10^(-4)M`
`:.pH=-logH^(+)=-log9.1xx10^(-4)`
`pH=3.04`