HX is a weak acid that reacts with water according to the following equation: Wh
ID: 811381 • Letter: H
Question
HX is a weak acid that reacts with water according to the following equation:
What is the equilibrium concentration of hydronium ion in a solution that is 0.0757 M in HX and 0.292 M in X1- ion?
Please show work, thanks!
HX(aq) + H2O(l) H3O1+(aq) + X1-(aq) Ka = 6.31e-06 HX is a weak acid that reacts with water according to the following equation: HX(aq) + H2O(l) H3O^1+(aq) + X^1-(aq) Ka = 6.31e-06 What is the equilibrium concentration of hydronium ion in a solution that is 0.0757 M in HX and 0.292 M in X^1- ion? Please show work, thanks!Explanation / Answer
This one is pretty easy since you have all data:
1) Remember that:
Keq = [product]/[reactants]
Ka = [H+][X-]/[HX]
Ka, [X-] and [HX] are given, you just need to solve for [H+]
Ka = [H+][X-]/[HX]
Ka*[HX]/[X-] = [H+]
substitute values:
NOTE: i will suppose 6.31e-06 is the same as 6.31x10^-06...
(6.31x10^-6)*(0.0757 M)/(0.292M) = [H+]
[H+] = 1.6358 x 10^-6
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