The lead(II) nitrate solution has a molarity of 1.2M if 22.00 mL of 2.00 M potassium iodide is required to equilibrate with 18.00 mL of lead(II) nitrate.
The following formula can be used to determine a solution's molarity:
na/nb = CaVa/CbVb
Where;
Ca is the acid concentration.
Cb = base concentration
Va = acid volume
Vb = base volume
na = number of acid moles
nb = number of base moles
The reaction's balanced equation is as follows:
2KNO3 (aq) + PbI2 Pb(NO3)2 (aq) + 2KI (s)
22 × 2/18 × Cb = 2/1
44/18Cb = 2
Cb = 44 ÷ 36
Cb = 1.2M
Since 22.00 mL of 2.00 M potassium iodide is required to equilibrate with 18.00 mL of lead (II) nitrate, the lead(II) nitrate solution's molarity is 1.2M.
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