The Ka of phenol, C6H5OH, is 1.00 x 10-10. What is [H3O+] in a 0.758 M solution
ID: 606507 • Letter: T
Question
The Ka of phenol, C6H5OH, is 1.00 x 10-10. What is [H3O+] in a 0.758 M solution of phenol? Use E notation and 3 significant figures.Explanation / Answer
C6H5OH + H2O ----------> C6H5O- + H3O+ .............................intial concentrations of C6H5OH = .758 M and C6H5O- = 0 M and H3O+ = 0 M ..............at equilibrium let [H3O+] = 'x' then [C6H5O-] = 'x' and [C6H5OH] = (.758-x) M and as Phenol is a weak acid x is very small compared to .758 M so we can write (.758-x) as .758M...........................and we have Ka = [H3O+][C6H5O-] / [C6H5OH]...................plugging the values we get 1.00*10-10 = x*x / .758............which gives x = 8.706*10-6 .........so [H3O+] = 8.706*10-6 M
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