%3Cp%20id%3D%22yui_3_11_0_24_1396902777866_113%22%20class%3D%22c15%22%3EAs%20a%2
ID: 828418 • Letter: #
Question
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/ Answer
The question is assuming that when acid is added to XAc (Ac = acetate ion), it reacts with the Ac forming X and HAc b/c H has a strong attraction for Ac which gives it the ability to force XAc to dissolve.
The equation for disassociation of all the ions involved is:
......................XAc(s) + HAc(aq) ~> X(aq) + H(aq) + Ac(aq)
initial conc......................1M ............. 0............ 0.......... 0
change.............................. -a............ 0.006........ a..... a+0.006
final................................. 1-a ............ 0.006....... a...... a+0.006
The acid-base equilibrium relation for acetic acid is used to find H and pH:
Ka = [H][Ac]/[HAc] = a(a+0.006)/(1-a) = 10^(-4.76) = 1.74x10^-5.
The 10^(-4.76) is the Ka obtained from pKa.
Simplify the equation: a^2+0.0060174a-1.74x10^-5=0
Solve: a=0.00213M=[H]
pH=-log[H]=2.67
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