The concentration required to begin to precipitate PbCl2 for PbCl2 is 0.0216 M.
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What is molarity?</h3>
Molarity is the measure of the concentration of any solute in per unit volume of the solution.
The reaction is 
The molarity of lead is 0.025 M
The ksp is given 17×10⁻⁵
Now, calculating the concentration
![[Pb^2^+] = 0.025 M.\\Ksp = 1.17 \times 10^-^5\\Ksp = [Pb^2^+] \times [Cl^-]^2\\[Cl^-] = \dfrac{\sqrt{ Ksp}}{[Pb^2^+]} \\\\[Cl^-] = \dfrac{\sqrt{ 0.0000117}}{0.025} \\[Cl^-] = 2.16 \times 10^-^2M.](https://tex.z-dn.net/?f=%5BPb%5E2%5E%2B%5D%20%3D%200.025%20M.%5C%5CKsp%20%3D%201.17%20%5Ctimes%2010%5E-%5E5%5C%5CKsp%20%3D%20%5BPb%5E2%5E%2B%5D%20%5Ctimes%20%20%5BCl%5E-%5D%5E2%5C%5C%5BCl%5E-%5D%20%3D%20%20%5Cdfrac%7B%5Csqrt%7B%20Ksp%7D%7D%7B%5BPb%5E2%5E%2B%5D%7D%20%5C%5C%5C%5C%5BCl%5E-%5D%20%3D%20%20%5Cdfrac%7B%5Csqrt%7B%200.0000117%7D%7D%7B0.025%7D%20%20%5C%5C%5BCl%5E-%5D%20%3D%202.16%20%5Ctimes%2010%5E-%5E2M.)
Thus, the concentration required to begin to precipitate PbCl2 for PbCl2 is 0.0216 M.
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A dissolved mixture when it mixed together hope this helps
From a stock solution of 3.00 m nitric acid, 9.391 ml of stock solution is needed to create a 0.161 m nitric acid solution, which has a total volume of 175 ml of the diluted solution.
A chemical reagent is present in vast quantities as a stock solution. It has a uniform concentration. Examples of typical stock solutions in laboratories are nitric acid and hydrochloric acid. These play a critical role in creating the titration-related solution preparations.
We know the formula for dilution type problems
M1 VI = M2 V 2
Where,
M, = initial molarity
V , = initial Volume
M2 = final molarity
V 2 = final Volume
Hene given -
M, = 3.00 M
VI = ?
M2 = 0.161M
V 2 = 175 ml
Accordingly ' MI V1 = M2 V 2
V1 =
V1= (0.161M*175ml)/ 3.00M
v1 = 9.391
The required volume of Stock solution is 9.391ml.
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