Use the given values of Ka to arrange the following acids in order of decreasing percent dissociation? Estimate [H3O+] in a 1.7 M solution of each acid. Having a difficult time trying to figure this out. Please help, Thank you.

Acid Formula Ka
Formic HCO2H 1.8×10−4
Hydrocyanic HCN 4.9×10−10
Perchloric HClO4 very large
Hypobromous HOBr 2.0×10−9

Part 1.
Use the given values of Ka to arrange the following acids in order of decreasing acid strength.
Use the given values of Ka to arrange the following acids in order of decreasing percent dissociation

Part 2.
Estimate [H3O+] in a 1.7 M solution of each acid.

[H3O+]formic, [H3O+]hydrocyanic, [H3O+]perchloric, [H3O+]hypobromous = ?

1 Answer
Nov 29, 2017

#K_a# measures the completion of the equilibrium.....

Explanation:

#HA(aq) + H_2O(l) rightleftharpoonsH_3O^+ +A^-#

#K_a=([H_3O^+][A^-])/([HA])#

And clearly, given this expression, STRONGER acids, i.e. those acids for which the equilibrium lies to the RIGHT as we face the page, have inherently high #K_a# values.

You have got #HClO_4#, #"perchloric acid"#, which is an exceptionally strong Bronsted acid, which would be stoichiometric in #H_3O^+# and #ClO_4^(-)#.

#underbrace(HCN, HOBr, HCO_2H, HClO_4)_rarr# #"increasing acid strength following the magnitude of"# #K_a#. And percent dissociation also increases in this order.

We will solve the equilibrium concentrations for formic acid (of course the strong acid is stoichiometric in #H_3O^+#)

#HOC(=O)H(aq) + H_2O(l) rightleftharpoonsH_3O^+ +H(O=)CO^-#

#K_"eq"=([H(O=)CO^(-)][H_3O^+])/([HOC(=O)H])#

...and if we let #[H_3O^+]=x#

#K_"eq"=x^2/(1.70-x)=1.80xx10^-4#...now this is a quadratic in #x#, which we could solve directly by the quadratic equation.... to make it easier, we make the approximation that #1.70-x~=1.70#

So #x_1=sqrt(1.70xx1.80xx10^-4)=0.0175*mol*L^-1#...and we recycle this estimate....and plug it in again....

#x_2=sqrt((1.70-0.0175)xx1.80xx10^-4)=0.0174*mol*L^-1#

So #[H_3O^+]=0.0174*mol*L^-1# for the formic acid solution. You have to repeat this calculation for the other weak acids....