To a 200 ml of 0.1M weak acid HA solution 90 ml of 0.1 molar solution of naoh be added now what volume of 0.1 molar naoh be added into above solution so that ph of a resulting solution be 5(Ka (HA)=10^-5) 1
a) 20ml
B) 2ml
C) 10ml
d) 15ml
ashu
12 Years agoGrade 11
2 Answers
Sunil Kumar FP
12 Years ago
ha + naoh---na+ + a- + h20
mm of ha=20(200×.1)
mm of naoh=9
mm of ha left=11
mm of a- formed=9
let the volume be v
mm of oh- added=.1×v
mm is the milimole
ph=pka +log[salt/acid]
5=5 + log[9+.1v/11-.1v]
v=15ml
thanks and regards
sunil kr
askIItian faculty
Sunil Kumar FP
12 Years ago
sorry while calculating answer will be 10 ml and not 15 ml.by mistake i wrote wrong