A 0.125 M solution of a weak acid is 2.5% ionized, what is the dissociation constant #K_c# of the acid?

Question

Reaction is
#HCN"(aq)"+H_2O"(l)"\rightleftharpoonsH_3O^{+}"(aq)"+CN^{-}"(aq)"#

Noah Aldrich · Accepted Answer

Regarding the response of a common weak acid to water, #"HA"(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "A"^(-)(aq)# If it is #2.5%# ionized, then the fraction of dissociation is #alpha = 0.025#. That is not #x#, whereas you have assumed that #x = alpha#. It cannot be true, because #x# is an absolute quantity, and #alpha# is a fraction, a relative quantity. Writing the traditional ICE table for any initial concentration #["HA"]_i#: #"HA"(aq) " "+" " "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "A"^(-)(aq)# #"I"" "["HA"]_i" "" "" "color(white)(..)-" "" "" "0" "" "" "" "" "0#
#"C"" "-x" "" "" "color(white)(....)-" "" "+x" "" "" "" "+x#
#"E"" "["HA"]_i-x" "color(white)(..)-" "" "" "x" "" "" "" "" "x# As a result, we established the standard mass action expression for monoprotic acid as: #K_c = (["H"_3"O"^(+)]["A"^(-)])/(["HA"]) = (xcdotx)/(["HA"]_i - x)# No approximations should be made here, because no #K_c# is known yet. All we need to do is define  #x = alpha cdot ["HA"]_i#, i.e. the extent of dissociation #x# can be written in terms of the initial concentration of #"HA"# and the fraction of dissociation #alpha#.  We just have a straightforward conversion from percentage to decimal, so nothing new here: #alpha = "percent ionization"/(100%)# Consequently, #K_c = (alpha["HA"]_i)^2/(["HA"]_i - alpha["HA"]_i)# #= (alpha^2["HA"]_i^2)/((1-alpha)["HA"]_i)# #= ((alpha^2)/(1-alpha))["HA"]_i# As a result, we already have enough information to find #K_c#. #color(blue)(K_c) = (0.025^2)/(1-0.025)(0.125)# #= color(blue)(8.0_128 xx 10^(-5))# where two significant figures are shown, with subscripts denoting digits past the final significant digit. Another strategy is to be aware of the ionization percentage, which is: "Percent ionization" is calculated as x/(["HA"]_i) xx 100%, or 2.5%. That's what you were initially doing, so #x = (2.5%)/(100%) xx ["HA"]_i = 0.025 cdot ["HA"]_i# #= ul(alpha cdot ["HA"]_i)# exactly as we previously defined. In this instance, we had #x = 0.025 cdot "0.125 M" = "0.003125 M"#.  From this not too dissimilar method... #color(blue)(K_c) = x^2/(0.125 - x)# #= (0.003125^2)/(0.125-0.003125) = color(blue)(8.0_128 xx 10^(-5))# We receive the same result.

Jose Ayala · Answer

undefined The dissociation constant $ K_c $ for a weak acid ($ HA $) is given by the expression:

$$ K_c = \frac{[H^+][A^-]}{[HA]} $$

Given that the solution is 2.5% ionized, this means that 2.5% of the weak acid has dissociated into $ H^+ $ ions and $ A^- $ ions.

Let's assume that the initial concentration of the weak acid $ HA $ is $ C $. Then, after dissociation, the concentration of $ H^+ $ ions and $ A^- $ ions will both be $ 0.025C $ (since they are 2.5% of the initial concentration).

The concentration of the undissociated weak acid, $ [HA] $, will be $ C - 0.025C = 0.975C $.

Substituting these values into the expression for $ K_c $, we get:

$$ K_c = \frac{(0.025C)(0.025C)}{0.975C} $$

$$ K_c = \frac{0.000625C^2}{0.975C} $$

$$ K_c = 0.000641C $$

Given that the initial concentration of the weak acid ($ C $) is 0.125 M, we can substitute this value into the expression for $ K_c $:

$$ K_c = 0.000641 \times 0.125 $$

$$ K_c = 8.0125 \times 10^{-5} $$

So, the dissociation constant $ K_c $ of the weak acid is $ 8.0125 \times 10^{-5} $.

A 0.125 M solution of a weak acid is 2.5% ionized, what is the dissociation constant #K_c# of the acid?

Reaction is #HCN"(aq)"+H_2O"(l)"\rightleftharpoonsH_3O^{+}"(aq)"+CN^{-}"(aq)"#

Reaction is
#HCN"(aq)"+H_2O"(l)"\rightleftharpoonsH_3O^{+}"(aq)"+CN^{-}"(aq)"#