If you know the [OH], how can you determine the pH of a solution? Match the words in the left column to the appropriate blanks in the sentences on the right. Reset Help 1.0 x 10“[H3O'] × [OH ] When the [OH ] is known, the [H3O+] can be calculated using the equation pH = – log [H3O*] which is called the 1.0 x 10 14 = [H30+]× [OH After rearranging to solve for [H3O+], this equation can then be substituted into the expression to solve for pH, which is water neutralization expression to determine the pH of a solution from the given [OH ]. water equilibrium equation pH =– log [OH ] 1.0 x 107 = [H3O+] × [OH] pH = log (H3O*] %3D water dissociation expression pH = 0.5 log [H3O*] %3D 1.0 x 10-7 = [H30*] × [OH ] %3D

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If you know the [OH ], how can you determine the pH of a solution?
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Match the words in the left column to the appropriate blanks in the sentences on the right.
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1.0 x 1014[H3O+]× [OH ]
When the [OH ] is known, the [H3O+] can be calculated using the equation
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pH = - log [H3O+]
which is called the
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>
1.0 x 10 14
[H3O+] x [OH
After rearranging to solve for [H30+], this equation can then be substituted into the expression to
water neutralization expression
solve for pH, which is
to determine the pH of a solution from the given [OH ].
water equilibrium equation
pH = - log [OH]
1.0 x 107 = [H3o*]× [OH]
pH = log [H3O*]
water dissociation expression
pH = 0.5 log [H30*]
|1.0 x 10-7= [H3O*] × [OH¯]
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Transcribed Image Text:Course Home Ô https://openvellum.ecollege.com/course.html?courseld=16516363&OpenVellumHMAC=6607995b61ef1ebc798fc86a6680d4f. O My Courses < Sections 10.5-10.7 Course Home Problem 10.38 - Enhanced - with Feedback > 8 of 17 Syllabus If you know the [OH ], how can you determine the pH of a solution? Scores Match the words in the left column to the appropriate blanks in the sentences on the right. eТеxt Reset Help Document Sharing 1.0 x 1014[H3O+]× [OH ] When the [OH ] is known, the [H3O+] can be calculated using the equation User Settings pH = - log [H3O+] which is called the Course Tools > 1.0 x 10 14 [H3O+] x [OH After rearranging to solve for [H30+], this equation can then be substituted into the expression to water neutralization expression solve for pH, which is to determine the pH of a solution from the given [OH ]. water equilibrium equation pH = - log [OH] 1.0 x 107 = [H3o*]× [OH] pH = log [H3O*] water dissociation expression pH = 0.5 log [H30*] |1.0 x 10-7= [H3O*] × [OH¯] 7:56 PM P Type here to search O O G 4)) 4/29/2021
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3.75 g of LiCl in 47.5 mL of LiCl solution
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Express your answer using three significant figures.
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6.0 g of casein in 120 mL of low-fat milk
Express your answer using two significant figures.
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Transcribed Image Text:Course Home 8 https://openvellum.ecollege.com/course.html?courseld=16516363&OpenVellumHMAC=6607995b61ef1ebc798fc86a6680d4f. Mastering Chemistry Course Home O My Courses < Section 9.4 Course Home Problem 9.40 - Enhanced - with Feedback 8 of 11 Syllabus 3.75 g of LiCl in 47.5 mL of LiCl solution Scores Express your answer using three significant figures. eТеxt V ΑΣφ ? Document Sharing |0.079 % User Settings Submit Previous Answers Request Answer Course Tools > X Incorrect; Try Again; 5 attempts remaining Part B 6.0 g of casein in 120 mL of low-fat milk Express your answer using two significant figures. Hνα ΑΣφ ? % 8:27 PM P Type here to search O O G 4) 4/29/2021
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