Question

what will be the ph of a solution obtained by mixing 800ml of 0.05M NaOH and 200 ml of 0.1M HCl ( assuming complete ionisation of the acid and base)​

Answer 1

MV =Milli moles of NaOH

=0.05x800=40

Milli moles of HCL =200x0.1=20

Milli moles Base more so resulting solution nature is basic.

OH- = (MV)base- (MV)acid/volume

(OH) - =40 -20/1000 =2/100= 0.02

pOH =-log (OH-)

POH =2-log2

POH = 2 -log0.3

POH =1.7

pH +POH =14

pH =14-1.7=12.3