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question mark

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

tejas sanjay akole , 16 Years ago
Grade 12
anser 1 Answers
askiitian.expert- chandra sekhar

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

Last Activity: 16 Years ago
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