A saturated solution of acid contains 11g/L and has pH = 2.94.calculate the Ka f

Question

A saturated solution of acid contains 11g/L and has pH = 2.94.calculate the Ka for the acid
HC6H11O2 (aq) + H2O(l) = H3O(aq ) + C6H11O2 (aq)
HC6H11O2 (aq) + H2O(l) = H3O(aq ) + C6H11O2 (aq)

Explanation / Answer

HA + H2O => H3O+    + A- molar mass of acid 116.2 g/mol 11g/L *(1 mol/116.2 g) = 0.0946643718 M [HA] + [A-]= 0.0946643718 M initially, all is HA             HA             + H2O => H3O+    + A- equil   0.0947 -x                        x     x x = [H3O+] = 10^-2.94 = 0.00114815362 Ka = [A-][H+]/[HA] = (0.00114815362)*(0.00114815362)/(0.0946643718-0.00114815362) Ka= 1.40965574e-5 ~ 1.41 e-5

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