1. What would be the concentration of A at equilibrium if the following reaction
ID: 978815 • Letter: 1
Question
1. What would be the concentration of A at equilibrium if the following reaction starts with 0.120M of A? 2A(aq) <==> B(aq) + C(aq) Kc=0.088
2.A solution of calcium hydroxide, Ca(OH)2 is created by dissolving 0.00489 grams of Ca(OH)2 in 69.5 mL of solution. Calculate the ph of this solution. Molar mass of Ca(OH)2 is 74.1 grams/mole.
3.Calculate the [H3O+] of a solution that has a pOH of 3.843
4.Determine the ph value of a solution at 25C containing 2.26 x 10^-4 [OH-] ions and state the acid/base nature of the solution.
5. Determine the ph of a 9.75 x 10^-2 M solution of Lactic Acid, HC3H5O3.
Explanation / Answer
1.
2A(aq) <==> B(aq) + C(aq) Kc=0.088
ICE table
[A]
[B]
[C]
initial
0.120 M
0
0
change
-x
+x
+x
equilibrium
0.120-x
x
x
Kc=[B][C]/[A]^2=x^2/(0.120-x)^2
0.088= x^2/(0.120-x)^2
Solving for x ,
X=0.123 M=[B]=[C]
[A]=0.12-x=0.12-0.12=0
2)Moles of Ca(OH)2=0.00489g/74.1g/mol=6.6*10^-5 moles
Molar concentration of Ca(OH)2=6.6*10^-5 moles/0.0695 L=94.96*10^-5 mol/L
Ca(OH)2-àCa2+ + 2OH-
[OH-]=2[Ca(OH)2]=2*94.96*10^-5 moles=189.92 *10^-5 mol/L
pOH=-log[OH-]=-log 189.92 *10^-5=2.72
pH=14-pOH=14-2.72=11.28
3. pOH = 3.843
pH=14-pOH=14-3.843=10.157
pH=-log[H3O+]
[H3O+]=10^-10.157=6.97*10^-11M
4
pOH=-log[OH-]=-log (2.26 *10^-4)=3.65
pH=14-3.65=10.35
5. HC3H5O3ß->H+ + [C3H5O3]-
[H+]=[ HC3H5O3]= 9.75 * 10^-2 M
pH=-log [H+]=-log(9.75 * 10^-2 M)=1.01
[A]
[B]
[C]
initial
0.120 M
0
0
change
-x
+x
+x
equilibrium
0.120-x
x
x
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