17 Question (2 points) At 25°C the Ksp for PbCl2 is 1.6 x...

17 Question (2 points) At 25°C the $K_{sp}$ for PbCl$_2$ is 1.6 x 10$^{-5}$. 1st attempt See page 800 Part 1 (1 pt) See Periodic Table Calculate Q for the following: 125.0 mL of 0.0300 M Pb(NO$_3$)$_2$ is mixed with 75.0 mL of 0.0200 M NaCl at 25°C Part 2 (1 pt) Will a precipitate form from the above reaction?

Transcript

00:01 Hello everyone let's look at the question so here now the question is at 25 degrees celsius the case for lead chloride is given and we have to calculate reaction quotient for the following okay now so the the given thing first let us write the given thing okay so the case p -sv value for lead chloride pb cl2 is 1 .6 multiplied by 10 is to minus 5 okay now the other thing that is given is so here molarity of polarity of pb 2 plus or the lead nitrate is equal to 0 .03 molar and molarity of nacl is equal to 0 .02 molar okay so let me write it as lead nitrate only and if we calculate the total volume of solution so after mixing lead nitrate with sodium chloride so the total volume of solution is so 125 plus 75 which is equal to so 200 m .m.

01:42 So after mixing these two the total volume becomes 200 ml right now so we have to write the reaction quotient.

01:51 Now see here when lead nitrate reacts with lead nitrate reacts with nacl so it forms lead chloride plus nano3.

02:07 Now lead chloride is the precipitate in this reaction now we have to write the reaction quotient for lead chloride because see here we can write the reaction question only for a reversible reaction okay now reaction quotient for lead chloride is pb cl 2 so this dissociates as pb 2 plus plus 2 cl minus now so we can write the reaction caution only when the this reversible reaction is not at equilibrium okay so this is equal to concentration of pb 2 plus multiplied by concentration of cl minus to the power 2 divided by concentration of pbcl 2 but so this p b, cl 2 is a solid, that's why, so it is, it will consider as 1.

02:57 So that's why it will be equal to concentration of pb 2 plus multiplied by concentration of cl minus to the power 2.

03:05 Okay.

03:06 Now this is the reaction question.

03:07 Now we have to find out the concentration of pb2s ions and cl minus ions...

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