Answer to Question #201435 in Physical Chemistry for madison

Question

Does a precipitate form if you add 15.00 𝑚𝐿 of a 2.10 × 10−3 𝑚𝑜𝑙/𝐿 solution of lead (II) nitrate to 25.00 𝑚𝐿 of a 2.50 × 10−2 𝑚𝑜𝑙/𝐿 solution of potassium chloride? Justify your answer mathematically. (𝐾𝑠𝑝 𝑓𝑜𝑟 𝑙𝑒𝑎𝑑(𝐼𝐼)𝑐h𝑙𝑜𝑟𝑖𝑑𝑒 𝑖𝑠 1.7 × 10−5)

Answer 1

Lead nitrate and potassium chloride reaction:

Pb(NO3)2(aq) + 2KCl(aq) + PbCl2(8) 1 + 2KNO3(aq)

Concentration of lead nitrate:

Molarity = 2.50 × 10−3 mol/L

Volume = 25.00 mL

1000 mL solution of lead nitrate has 2.50 × 10−3 moles

So, 25.00 mL solution of lead nitrate has (2.50 × 10−3 / 1000) x 25 = 62.5 x 10-6 = 6.25 x 10-5 moles.