Will a precipitate form when 20.0 ml of 0.10 M Ba(NOxaq) and 50.0 mL of 0.10 M NaCO(aq) are mixed together? A. Q < K precipitate will form
B. Q > K precipitate will form
C. Q > K precipitate will not form
D. Q < K precipitate will not form
E. Cannot determine. Need more information

B. Q > K precipitate will form

Explanation:

The reaction is;

Ba(NO3)2(aq) + Na2CO3(aq) > BaCO3(s) + 2NaNO3(aq)

Hence the reaction could form a precipitate of BaCO3.

Number of moles of carbonate ions = 50/1000 * 0.10 M =  5 * 10^-3 moles

Number of moles of Barium ions = 20/1000 * 0.10 M = 2 * 10^-3 moles

Total volume after reaction = 20ml + 50ml = 70 ml or 0.07 L

Molarity Barium ions = 5 * 10^-3 moles/ 0.07 L = 0.07 M

Molarity carbonate ions = 2 * 10^-3 moles/ 0.07 L =0.03 M

Q = [Ba^2+] [CO3^2-] = 0.07 * 0.03 = 2.1 * 10^-3

But K = 2.58  ×  10 ^− 9

We can clearly see that Q>K therefore precipitate will form

pure substance: iron,copper

impure substance: water,charcoal

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