Which answer best describes what happens when 750 mL of 4.00 x 10 -3 M Ce(NO 3 )

Question

Which answer best describes what happens when 750 mL of 4.00 x 10-3 M Ce(NO3)3 is added to 300.0 mL of 2.00 x 10-2 M KIO3? Ksp = 1.9 x 10-10 for Ce(IO3)3. <?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

(A) Qsp = 5.3 x 10-10; precipitate will not form                (B) Qsp = 1.6 x 10-5; precipitate will not form

(C) Qsp = 5.3 x 10-10; precipitate will form                       (D) Qsp = 1.6 x 10-5; precipitate will form

(E) Qsp = 8.0 x 10-5; precipitate will form

ANSWER IS C. SOMEONE PLEASE HELP EXPLAIN IN FULL DETAIL. I HAVE A TEST TODAY

Explanation / Answer

[Ce(NO3)3] = 750/1050 * 4*10^-3 M = 2.857*10^-3 M

[KIO3] = 300/1050 * 2.00*10^-2 M = 5.714*10*-3 M

for Ce(IO3)3, Qsp = [IO3-]

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