146. 50 ml of 0.2N H.SO, is mixed with 100 ml of
0.4N KOH solution and 1.85 lit of distilled water
is added. The pH of resulting solution is
(log 1.5 = 0.176)
1) 13.301 2) 0.699
3) 1.824
4) 12.176
Answers
Answered by
4
Answer:
12.176
Explanation:
normality of H2SO4 = 0.2N
normality of KOH = 0.4N
volume of H2SO4 = 50ml
volume of KOH = 100 ml
volume of water = 1.85 L = 1850ml
normality of water = 0N
Nr = (N1V1~N2V2~N3V3)/V1+V2+V3
Nr = (50×0.2 ~ 100×0.4 ~ 1850×0)/50+100+1850
Nr = 10~40~0/ 2000
Nr = 30/2000
Nr = 1.5×10^-²
since NbVb > NaVa the solution is basic
pOH = -log(1.5×10^-²)
pOH = 2-0.176
pH+pOH = 14
pH = 14-pOH
pH = 14-(2-0.176)
pH = 14-2+0.176
pH = 12+0.176 = 12.176
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