Homework Answers
Zn2+ + 4OH- ---------> Zn(OH)42-
Kc = [Zn(OH)42-]/[Zn2+][OH-]4
Initial concentration
[Zn2+] = 0.10M/(35ml/25ml) = 0.0714M
[OH-] = 0.65M/(35ml/10ml) = 0.1857M
[Zn(OH)42-] = 0
Change in concentration
[Zn2+] = -x
[OH-] = -4x
[Zn(OH)42-] = + x
equillibrium concentration
[Zn2+] = 0.0714 - x
[OH-] = 0.1857 - 4x
[Zn(OH)42- ] = x
pH = 9.79
pOH = 14 - 9.79 = 4.21
pOH = -log[OH-]
-log[OH-] = 4.21
[OH-] = 6.17 × 10-5M
so,
0.1857 - 4x = 6.17 ×10-5
4x = 0.1856
x = 0.0464
so, at equillibrium
[Zn+] = 0.0714 - 0.0464 = 0.025M
[Zn(OH)42-] = 0.0464M
substituting equillibrium concentration values
Kc = 0.0464M/((0.025M)(0.0000617)4
Kc = 1.28×1017
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REPOSTING: observing pH changes in Water and Buffer
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the amount added of HCl and NaOH are both 25mL.
the buffer solution is made of 2grams of NaC2H3O2 and
4 mL of 6M HC2H3O2 in a 50 mL solution (46mL of water). the
solution will contain 2.4x10^-2 mol each of NaC2H3O2 a d
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how to answer it (I'm not sure I'm doing it right and
want an expert to double check):
I need help calculating the theoretical and...
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