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Grade 11Physical Chemistry

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

Profile image of ashu
12 Years agoGrade 11
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2 Answers

Profile image of 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

Profile image of Sunil Kumar FP
12 Years ago
sorry while calculating answer will be 10 ml and not 15 ml.by mistake i wrote wrong

thanks and regards
sunil kr
askIItian faculty