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calculate[H+] in solution that is 0.1M HCOOH and 0.1M HOCN
Ka(HCOOH)=1.8*10-4& Ka(HOCN)=3.3*10-4
Hi Tejas,
[H+][HCOO-]/[HCOOH]=1.8*10-4
(x+y)*x/(1-x)=1.8*10-4------(1)
[H+][OCN-]/[HOCN]=3.3*10-4
(x+y)*y/(1-y)=3.3*10-4-------(2)
(1)/(2)=x/y=1.8/3.3 considering 1-x≈1 and 1-y≈1
y=3.3*x/1.8
substituting in (1)
5.1/1.8 *x2 = 1.8*10-4
x=7.97*10-3
y=14.612*10-3
x+y=22.582*10-3
[H+]= 0.0226
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