Question

chem help pls

Answer 1

Hi, welcome to QC, I would love to help you. What do you need help with?

Answer 2

i can dm

Answer 3

you cant put it on here?

Answer 4

hold on let me try to attach the file

Answer 5

@AZ

do you know chemistry?

Answer 6

I can try to help you

Answer 7

Let's do one at a time

For \(\sf AgCl\), the \(K_{sp}\) is \( 1.8\times 10^{-10}\)

And you have correctly written out the \(\sf K_{sp}\) expression in part A.

We also know that the solution has \(\sf 0.10~ M~~ AgNO_3\)

so that means the concentration of \(\sf Ag^+\) is \(\sf 0.10 ~M\)

so plug it into your \(\sf K_{sp}\) expression and solve for the concentration of \(\sf Cl^-\) needed to precipitate \(\sf AgCl\)

Answer 8

For \(\sf AgCl\)

\(\sf K_{sp} = [Ag^+]~[Cl^-]\)

and we know that
\(\sf K_{sp}~ for ~AgCl = 1.8\times 10^{-10}\)

and that
\(\sf [Ag^+] = 0.10~M\)

so plug it in and solve for \(\sf [Cl^-]\)

all you have to do is divide on both sides by 0.10

Answer 9

And now for \(\sf PbCl_2\)

\(\sf K_{sp} = [Pb^{+}]~[Cl^-]^2\)

and we know that
\(\sf K_{sp}~ for ~PbCl_2 = 1.6\times 10^{-5}\)

and that
\(\sf [Pb_2] = 0.20~M\)

so plug it in and solve for \(\sf [Cl^-]\)

all you have to do is divide on both sides by 0.20

Answer 10

And for part C)

The precipitate (\(\text{AgCl}\) or \(\text{PbCl}_2\)) that requires a \(\bf lower\) concentration of \(\sf [Cl^-]\) will be the one that precipitates \(\bf first\) since \(\sf Q_{sp} > K_{sp}\)