Question

# 20. Calculate the molarity of hydroxide ion in an aqueous solution that has a pOH of...

20. Calculate the molarity of hydroxide ion in an aqueous solution that has a pOH of 5.00

[A] 1.0 ⋅ 10-9 [B] 1.0 ⋅ 10-5 [C] 9.0 ⋅ 10-14 [D] 9.00 [E] 5.0 ⋅10-14

21. What is the pOH of an aqueous solution at 25.0 °C that contains 3.98 ⋅ 10-9 M hydronium ion?

[A] 9.000 [B] 5.600 [C] 3.980 [D] 7.000 [E] 8.400

22. Which solution will be the most basic?

[A] 0.20 M Sr(OH)2 [B] 0.20 M NH3 [C] 0.20 M NaOH [D] 0.20 M H2O [E] All solutions have equal basicity.

23. A 3.0 x 10-3 M aqueous solution of Ca(OH)2 at 25.0 °C has a pH of _______

[A] 11.78 [B] 6.0 x 10-3 [C] 2.22 [D] 11.48 [E] 1.7 x 10-12

20) pOH = 5.0 ; pOH = -log(OH-)
[OH]- = 10-5 M

21) M= 3.98 x 10-9

pOH = - log(3.98 x 10-9 )

pOh = 8.4

22) pH + pOH = 14 For a strong base pH is very high which means value of pOH must be very small

Sr(OH)2 disassociates into two OH- ions so [OH]- = 0.4 M

pOH = -log(0.4) = 0.397

For NH3 NH3 + H2O ----> NH4OH NH4+ + OH- Kb = 1.8x10-5

[OH]- = sqrt(KbxC) = 1.897 x 10-3 ; pOH = -log(1.897 x10-3) = 2.72

NaOH is ionic so it disassociates completely pOH = -log[0.2] = 0.698

As Sr(OH)2 has the least pOH value it is the most basic

23) Ca(OH)2 is a strong base so it disassocites into 2 OH- ions and one Ca2+ ion

[OH]- = 6 x 10-3

pOH = -log(6 x 10-3) = 2.221

pH + pOH = 14 ; pH = 14 - pOH = 14 - 2.221 = 11.779

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