Problem 16.45 Part A Calculate [OH] for 1.3×10 3 M Sr(OH)2. Express your answer
ID: 879576 • Letter: P
Question
Problem 16.45
Part A
Calculate [OH] for 1.3×103 M Sr(OH)2.
Express your answer using two significant figures.
SubmitMy AnswersGive Up
Part B
Calculate pH for 1.3×103 M Sr(OH)2.
Express your answer using two decimal places.
SubmitMy AnswersGive Up
Part C
Calculate [OH] for 2.260 g of LiOH in 280.0 mL of solution.
Express your answer using four significant figures.
SubmitMy AnswersGive Up
Part D
Calculate pH for 2.260 g of LiOH in 280.0 mL of solution.
Express your answer using four decimal places.
SubmitMy AnswersGive Up
[OH] = MExplanation / Answer
Part A -
[Sr(OH)2] = 1.3×103 M
As Sr(OH)2 is a strong base it will completely dissociate
Therefore, the hydroxide ion concentration in moles per liter, [OH-], will be twice the concentration of the Sr(OH)2.
[OH-] = 2*[Sr(OH)2]
= 2*1.3×103 M
= 2.6×103 M
Part B -
[Sr(OH)2] = 1.3×103 M
[OH-] = 2.6×103 M
pOH = - log ( 2.6×103)
= 2.58
pH = 14 - pOH
= 14 - 2.58
= 11.42
Part C -
2.260 g of LiOH in 280.0 mL of solution.
moles of LiOH = 2.260 g / 23.95 g/mol
= 0.0943
[LiOH] = 0.0943 / 0.280
= 0.336 M
As LiOH is a strong base, it will completely dissociate in the solution.
[OH-] = 0.336 M
Part D -
pH for 2.260 g of LiOH in 280.0 mL of solution.
[LiOH] = 0.336 M
[OH-] = 0.336 M
pOH = - log 0.336
= 0.473
pH = 14 - pOH
= 14 - 0.473
= 13.527