If 22.00 mL of 2.00 M potassium iodide is needed to reach the equivalence point with 18.00 mL of lead (II) nitrate, determine the molarity of the lead (II) nitrate solution. Note: First write the balance equation between potassium iodide and lead (II) nitrate.

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29-03-2022
Chemistry

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If 22.00 mL of 2.00 M potassium iodide is needed to reach the equivalence point with 18.00 mL of lead (II) nitrate, determine the molarity of the lead (II) nitrate solution. Note: First write the balance equation between potassium iodide and lead (II) nitrate.

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samueladesida43 samueladesida43 · Answer 1 · 2022-03-30T10:27:41+02:00

The molarity of the lead(II) nitrate solution if 22.00 mL of 2.00 M potassium iodide is needed to reach the equivalence point with 18.00 mL of lead (II) nitrate is 1.2M.

How to calculate molarity?

The molarity of a solution can be calculated using the following formula:

CaVa/CbVb = na/nb

Where;

Ca = concentration of acid
Cb = concentration of base
Va = volume of acid
Vb = volume of base
na = number of moles of acid
nb = number of moles of base

The balanced equation of the reaction is as follows:

Pb(NO3)2 (aq) + 2KI (aq) → 2KNO3 (aq) + PbI2 (s)

22 × 2/18 × Cb = 2/1

44/18Cb = 2

Cb = 44 ÷ 36

Cb = 1.2M

Therefore, the molarity of the lead(II) nitrate solution if 22.00 mL of 2.00 M potassium iodide is needed to reach the equivalence point with 18.00 mL of lead (II) nitrate is 1.2M.

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