Question

Please make sure the answer is in the correct sigfigs!

Suppose 6.34 g of lead(II) acetate is dissolved in 300. mL of a 0.40 Maqueous solution of ammonium sulfate. Calculate the final molarity of lead(II) cation in the solution. You can assume the volume of the solution doesn't change when the lead(II) acetate is dissolved in it. Be sure your answer has the correct number of significant digits. x 6 ?

Answer 1

Pb(CH3COO)2 + (NH4)2SO4 ----> 2NH4(CH3COO) + PbSO4

Given mass of lead(ii) acetate = 6.34 g

So, moles of lead(ii) acetate = 0.01949 mol

Number of moles of ammonium sulphate = 0.40 * 0.300 = 0.12 mol

Since 1 mol of lead(ii) acetate reacts with 1 mol of ammonium sulphate to give 2 mol of ammonium acetate and 1 mol lead sulphate.

Here limiting reagent is lead(ii) acetate.

So, moles of lead(ii) cation = moles of lead sulphate = moles of lead(II) acetate = 0.01949 mol

Volume of solution = 0.300 L

So, [Pb+2] = 0.065 M

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