Finding moles of Pb2+ from a solution of Pb(NO3)2

Question

When 25.0mL of 0.05M Pb(NO₃)₂ are mixed with 35.0mL of 0.01M KI, a yellow precipitate of PbI₂(s) forms.

a. How many moles of Pb²⁺ are initially present?

b. How many moles of I^- are initially present?

Answer 1

moles Pb2+ initially present:

Pb(NO₃)₂ (aq) ==> Pb²⁺ (aq) + 2NO₃^-(aq)

25 ml x 1 L/1000 ml x 0.05 mol/L = 0.00125 moles Pb²⁺ initially present

moles I- initially present:

KI (aq) ==> K⁺(aq) + I^-(aq)

35 ml x 1 L / 1000 ml x 0.01 mol/L = 0.0035 moles I^- initially present