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KHW27 Tro 14.9-14.10 (90 min) Exercise 14.108 Term 2: Significant Figures Feedback: Your answer 3.IU2 was either rounded differentiy number of significant figures than required for this part. Constants Perniodic Table For each strong base solution, determine OH HO] pH, and pOH PartG 3.5x10-4 M Ca(OH), determine [OH l and [H30] Express your answers using two significant figures. Enter your answers numerically separat [OH ] [H307 = 13.5 . I 0 4,2.9. 10-11 Submit Previous Answers Request Answer Incorrect; Try Again; 4 attempts remaining

wanianhips C Home KHW27 Tro 14.9-14.10 (90 min) Calculating [H30+] and [OH-] In Solutions It may help to keep the following equation in mind as you work K, = [H30+ IOH |-|1 × 10-7 M[1 × 10-7 M = 1 × 10-14 M Drag the appropriate items to their respective bins View Available Hint(s) Reset Help Basic Acidic Neutral

KHW27 Tro 14.9-14.10 (90 min) Calculating [H30+] and [OH-] In Solutions A solution has HO-7.6x10-5 M. Use the ion product constant of water to find the OH of the solution. Express your answer to two significant figures. View Available Hint(s) Submit

Part C A solution has (OH] 1.5x10-2 M. Use the ion product constant of water to find the H30 of the solution. Express your answer to two significant figures. View Available Hintís) 图] ? H30] Submit

plz help on this ...

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Answer #1

1. Given that [Ca(OH)2] = 3.5 * 10-4 M

1 mol of Ca(OH)2 contains 2 mol of OH-

So,[OH-] = 2*3.5 * 10-4 M = 7.0 *10-4 M

Since , Kw = [H3O+][OH-] = 10-14

[H3O+] =  10-14 /  7.0 *10-4 M = 1.43 * 10-11 M

2. Since, , Kw = [H3O+][OH-] = 10-14

pH = -log[H3O+]

Now, If pH > 7 then solution is acidic   

= 7 neutral

< 7 . basic

[OH-] [H3o+] pH
6 * 10-12 0.00167 2.78 Acidic
3*10-9 3.33 * 10-6 5.48 acidic
2*10-10 5 * 10-5 4.30 acidic
4 * 10-13 0.025 1.60 acidic
4 * 10-2 3 * 10-13 12.52 Basic
7 * 10-4 1.43 * 10-11 10.85 Basic
6 * 10-5 1.67 * 10-10 9.78 Basic
1* 10-7 1* 10-7 7 Neutral

3. [H3O+] = 7.6*10-5

Kw = [H3O+][OH-] = 10-14

[OH-] = 10-14/ ( 7.6*10-5) = 1.32 * 10-10 M

4.[OH-] = 1.5 *10-2 M

Kw = [H3O+][OH-] = 10-14

[H3O+] =  10-14 / ( 1.5 *10-2 ) = 6.67 * 10-13

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