42.7 mL of aqueous 0.255 M Pb(NO3)

Question

42.7 mL of aqueous 0.255 M Pb(NO3)2 is mixed with 21.7 of mL of 0..415 M NaCl. The equation for the precipitate reaction is Pb(NO3)2 (aq) + 2NaCl(aq)—->PbCl 2 (s)+2 NaNO 3 (aq) The concentration of NO3^- ion in the reaction solution is what M.

Answer 1

Pb(NO3)2 (aq) + 2NaCl(aq)—->PbCl 2 (s)+2 NaNO 3 (aq)

NO3- is dissolved the whole time. The only thing that changes is the volume.

Molarity NO3- = ^{mol NO3-} / _{L solution}

How many moles of NO3- are there?

Pb(NO3)2 (aq)

^42.7mL/₁ * ^{0.255molPb(NO3)2}/_1000mL = 0.01089 mol Pb(NO3)2

0.01089 mol Pb(NO3)2 * ^{2mol NO3-}/_1molPb(NO3)2=0.0218 mol NO3-

What's the new volume of solution in Liters?

42.7mL+21.7mL=64.4mL

64.4mL/1000=0.0644L

Molarity NO3- = ^{mol NO3-} / _{L solution}

Molarity NO3- = ^0.0218mol/_0.0644L

ANSWER= 0.339 M