Exercise 15.81 For each of the following strong base solutions, determine [OH],[

Question

Exercise 15.81 For each of the following strong base solutions, determine [OH],[H3O+], pH, and pOH.

Part A 0.15 M NaOH Express your answer using two significant figures. Enter your answers numerically separated by commas. [OH], [H3O+] = M SubmitMy AnswersGive Up

Part B Express your answer to two decimal places. Enter your answers numerically separated by commas. pH, pOH = SubmitMy AnswersGive Up

Part C 1.8×103 M Ca(OH)2 Express your answer using two significant figures. Enter your answers numerically separated by commas. [OH], [H3O+] = M SubmitMy AnswersGive Up

Part D Express your answer to two decimal places. Enter your answers numerically separated by commas. pH, pOH = SubmitMy AnswersGive Up

Part E 5.0×104 M Sr(OH)2 Express your answer using two significant figures. Enter your answers numerically separated by commas. [OH], [H3O+] = M SubmitMy AnswersGive Up

Part F Express your answer to two decimal places. Enter your answers numerically separated by commas. pH, pOH = SubmitMy AnswersGive Up

Part G 8.9×105 M KOH Express your answer using two significant figures. Enter your answers numerically separated by commas. [OH], [H3O+] = M SubmitMy AnswersGive Up

Part H Express your answer to two decimal places. Enter your answers numerically separated by commas. pH, pOH = SubmitMy AnswersGive Up

Explanation / Answer

ANSWER

Dear candidate as per guidelines one question with meximum of 4 parts can be solved at one time, so here we solve first four parts please send others separaely.

Part(A) 0.15M NaOH

NaOH --------> Na+ + OH-

[OH-] = 0.15 Because NaOH dissociates completely.

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

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

[H3O+] = 10-14 / 0.15 = 6.66 X 10-14 M

Part (B)

pOH = -log[OH-] = -log[0.15] = 0.82

pH + pOH = 14

pH = 14 - pOH = 14 - 0.82 = 13.18

Part (C) 1.8×103 M Ca(OH)2

Ca(OH)2 -----------> Ca2+ + 2OH-

one mole of Ca(OH)2 gives two moles of OH-

1.8×103 M Ca(OH)2 will give 2 X1.8×103 M OH-

[OH-] = 2 X1.8×103 M = 3.6 ×103 M,

[H3O+] = 10-14 / 3.6 ×103 M = 2.7X 10-12

Part (D)

pOH = - log[OH-] = - log (3.6 ×103 ) = 2.44

pH = 14 - 2.44 = 11.56

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