Question

Can I get Chemistry help?

Answer 1

@AZ

Answer 2

oof I would help but idk Chemistry

Answer 3

so for MX2, our solubility product constant would be

\(\sf K_{sp} = [M][X]^2\)

we know the concentration of MX2 is 0.012 M

so the concentration of M is 0.012 and the concentration of X is going to be 2* 0.012 because we have 2 X for every one M

Answer 4

so you plug it in and solve for Ksp

Answer 5

so 6.9*10^-6

Answer 6

Yup!

Answer 7

For the second one, can you write the equation for the Ksp for Zn(OH)2

Answer 8

5.0x10-17.

Answer 9

What is that?

We want to write the Ksp equation first

Answer 10

\(\sf ZnOH2 --> Zn^{2+} + 2OH^{-}\)

What is the Ksp equation

Answer 11

Then you can use the Ksp to find the concentration of OH-

and remember that

[H+][OH-] = 10^(-14)

and then you can calculate the concentration of H+

and

pH = -log[H+]

Answer 12

3.00 x 10^-17.

Answer 13

is the ksp right?

Answer 14

no

they told you that the Ksp is 4.5 * 10^-17

It's in the question

we need to find the concentration of OH- from that

Answer 15

You first need to write the solubility product expression of \(\sf Zn(OH)_2\)

Answer 16

3.0*10^16

Answer 17

What is that???

Answer 18

You need to write the solubility product expression

Answer 19

i'm just lost. i apologize, this is a confusing problem to me

Answer 20

Zn(OH)2(s)-->Zn2++2OH-

Answer 21

Yes, and so this would be the Ksp expression right?

\(\sf\Large K_{sp} = [Zn^{2+}] [OH^{-}]^2\)

Answer 22

Yes

Answer 23

and so we know that the Ksp is 4.5 *10^-17 because they told us that in the question

\(\sf\Large 4.5 \times 10^{-17}= [Zn^{2+}] [OH^{-}]^2\)

and so if the concentration of Zn2+ is 'x'

then the concentration of OH- is going to be 2x

because there's 2 moles of OH- for every mole of Zn2+

(this is really similar to the previous question we just answered)

So first solve for 'x' but remember that we want to ultimately get the concentration of OH- so we're looking for 2x

\(\sf4.5 \times 10^{-17}= (x)(2x)^2\)

Answer 24

x is 2.24*10^6?

Answer 25

10^-6

then yes

Answer 26

yes

Answer 27

That would be the concentration of \( \sf Zn^{2+}\)

multiply that number by 2 to get \(\sf [OH^{-}]\)

Answer 28

4.48*10^-6

Answer 29

now remember

[H+] * [OH-] = 10^-14

so

we know [OH-], we just have to find the concentration of H+

\( \sf [H^{+}] \times 4.48 \times 10^{-6} = 10^{-14}\)

[H+] = ??

Answer 30

2.23*10^-9

Answer 31

Now, the last step is to remember how to calculate pH

pH = -log([H+])

so the pH = -log(2.23*10^-9)

pH = ??

Answer 32

8.17

Answer 33

Check your calculations again

https://www.google.com/search?q=-log(2.23*10%5E-9)

Answer 34

Thank you!

Answer 35

8.65***

Answer 36

my apologies

Answer 37

could you just check over some word problems now for me?

Answer 38

@AZ