Question

Calculate the ( mathrm{K}_{mathrm{sp}} ) of ( mathrm{Ca}(mathrm{OH})_{2} ) : 1. ( mathrm{pH}+mathrm{pOH}=14 ) 2. In dissolution of ( mathrm{Ca}(mathrm{OH})_{2},left[mathrm{Ca}^{2+} ight]=1 / 2[mathrm{OH}] ), find ( left[mathrm{Ca}^{2+} ight] ) 3. Calculate ( mathrm{K}_{mathrm{sp}} ) of ( mathrm{Ca}(mathrm{OH})_{2} ), by its definition, ( mathrm{K}_{mathrm{sp}}=left[mathrm{Ca}^{2+} ight][mathrm{OH}]^{2} ) Calculation: the ( mathrm{K}_{mathrm{sp}} ) of ( mathrm{Ca}(mathrm{OH})_{2} ) : (show all work) 1. ( [mathrm{OH}^{-}] = ) 2. ( left[mathrm{Ca}^{2+} ight] = ) 3. ( K_{mathrm{sp}} = )

          Calculate the ( mathrm{K}_{mathrm{sp}} ) of ( mathrm{Ca}(mathrm{OH})_{2} ) :
1. ( mathrm{pH}+mathrm{pOH}=14 )
2. In dissolution of ( mathrm{Ca}(mathrm{OH})_{2},left[mathrm{Ca}^{2+}
ight]=1 / 2[mathrm{OH}] ), find ( left[mathrm{Ca}^{2+}
ight] )
3. Calculate ( mathrm{K}_{mathrm{sp}} ) of ( mathrm{Ca}(mathrm{OH})_{2} ), by its definition, ( mathrm{K}_{mathrm{sp}}=left[mathrm{Ca}^{2+}
ight][mathrm{OH}]^{2} )

Calculation: the ( mathrm{K}_{mathrm{sp}} ) of ( mathrm{Ca}(mathrm{OH})_{2} ) : (show all work)
1. ( [mathrm{OH}^{-}] = )
2. ( left[mathrm{Ca}^{2+}
ight] = )
3. ( K_{mathrm{sp}} = )

Show more…

Calculate the ( mathrmKmathrmsp ) of ( mathrmCa(mathrmOH)2 ) :
1. ( mathrmpH+mathrmpOH=14 )
2. In dissolution of ( mathrmCa(mathrmOH)2,left[mathrmCa^2+
ight]=1 / 2[mathrmOH] ), find ( left[mathrmCa^2+
ight] )
3. Calculate ( mathrmKmathrmsp ) of ( mathrmCa(mathrmOH)2 ), by its definition, ( mathrmKmathrmsp=left[mathrmCa^2+
ight][mathrmOH]^2 )

Calculation: the ( mathrmKmathrmsp ) of ( mathrmCa(mathrmOH)2 ) : (show all work)
1. ( [mathrmOH^-] = )
2. ( left[mathrmCa^2+
ight] = )
3. ( Kmathrmsp = )

Added by Ayarpi R.

Chemistry and Chemical Reactivity

John C. Kotz, Paul M. Treichel, John R.… 10th Edition

Chapter 17

Instant Answer

Solved by Expert Jennifer Hudspeth

04/21/2024

Step 1

9, we can calculate the pOH using the formula: \[ \text{pH} + \text{pOH} = 14 \] \[ 11.9 + \text{pOH} = 14 \] \[ \text{pOH} = 14 - 11.9 = 2.1 \] Show more…

Show all steps

lock

Thanks for your feedback!

Please complete the worksheet given the PH is 11.9

Play audio

Feedback

Powered by NumerAI

Jennifer Hudspeth and 91 other subject Chemistry 102 educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

00:02 There, in this question, we are told that the ph is 11 .9.

00:06 And the first thing we want to do is find the hydroxide ion concentration.

00:11 Well, to do that, the first thing i want to know is what the poh is.

00:14 And we're told there in the question that ph plus poh add up to 14.

00:24 Therefore, since the ph is 11 .9, the poh is going to be 14 minus 11 .9.

00:34 Or in other words, the poh is going to be 2 .1.

00:40 Okay, so from that, we need to find the hydroxide ion concentration.

00:46 Well, there's another equation that is not given up to us here, but it tells us that the hydroxide ion concentration is going to be the inverse log or antilog of the negative poh.

01:03 And this just comes from the equation that tells us poh is the negative log of the hydroxide ion concentration.

01:12 So rewriting that equation, we get this.

01:15 So let's go ahead and calculate that oh-.

01:18 We need to take the anti -log, or your calculator may just call it the inverse log, of negative, the poh is 2 .1.

01:30 Okay, so we type in negative 2 .1.

01:32 We take the inverse log or anti -log of that, and we'll get our oh - concentration.

01:40 That oh - concentration is 7 .94 times 10 to the negative third.

01:53 This is the concentration, so it is molarity...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later