Calculate the pH of the resultant mixtures :
10 mL 0.1 M `H_2SO_4` + 10 mL 0.1 M KOH

Explanation

Millimol=mL x M
In Initial mol of `H_2SO_4` = 10 x 0.1 = 1.0 mol `H_2SO_4`
In initial m mole of KOH = 10 x 0.1 = 1.0 m mol KOH
Total volume of mix solution 20 mL = 0.02 L
Millimol of remaining `H_2SO_4` after neutralization
`{:(H_2SO_4 + , 2KOH to , K_2SO_4 + , 2H_2O),("1 mol","2 mol" , "1 mol", "2 mol"):}`
`therefore` (m mol of remaining `H_2SO_4` ) = (m mol `H_2SO_4` in initial ) -(`H_2SO_4` consumed in neutralization )
=(1 m mol - `1/2` m mol ) `H_2SO_4`
=0.5 m mol `H_2SO_4` remaining
=0.5/1000 mol `H_2SO_4` remaining
Molarity of remaining ,
`H_2SO_4 =("mol"/"Total volume (L)")`
`=(0.5xx1000)/(1000xx20)`
`=0.5/20`=0.025 M `H_2SO_4`
Remaining `[H^+]` in mixture = `2[H_2SO_4]`
=2(0.025) = 0.05
pH = log `[H^+]` = log (0.05)
=-(-1.3010)=1.3010