Calculate the `pH` of a solution which contains `100mL` of `0.1 M HC1` and `9.9 mL` of `1.0 M NaOH`.

Answer

C, D

Explanation

`{:(,HC1+,NaOHrarr,NaC1+,H_(2)O),("mEq before reaction",100xx0.1,9.9xx1,,),(,=0,=9.9,,),("mEq after reaction",0.1,0,9.9,9.9):}`
`:. [H^(o+)]` left from `HC1 =(0.01)/(109.9) = 9.099 xx 10^(-4)M`
`:. pH =- log [H^(o+)] =- log (9.099 xx 10^(-4))`
`pH = 3.0409`