Question

26. + -10.1 points 0/4 Submissions Used Calculate [OH-] and pH for each of the following solutions. (a) 0.0054 M CSOH [OH-] = M pH = (b) 0.0136 g of KOH in 530.0 mL of solution [OH-] = M pH = (C) 15.2 mL of 0.00283 M Sr(OH)2 diluted to 1000 mL [OH-] = M pH = (d) A solution formed by mixing 81.0 mL of 0.000470 M Sr(OH)2 with 38.0 mL of 5.6 x 10-3 M CSOH [OH-] = M pH =

Answer 1

a) CsOH is stong base dissociates completely

so [CsOH] = [OH^-] = 0.0054 M

[OH^-] = 0.0054 M

pOH = - log [OH^-]

pOH = - log [0.0054]

pOH =2.27

pH = 14 - 2.27

pH = 11.73

b) first calculate molarity of the solution

Molarity = (W /MW) (1000 / V in mL)

Molarity = (0.0136 / 56.1)(1000 / 530)

Molarity = 0.00046 M

as KOH is strong base

[KOH] = [OH^-] = 0.00046 M

[OH^-] = 0.00046 M

pOH = log [OH]

pOH = - log [0.00046]

pOH = 3.34

pH = 14 - 3.34

pH = 10.66

c) Sr(OH)2 ---------> Sr⁺² + 2OH^-

according to dilution law

M1V1 = M2V2

M2 = (M1V1/V2)

M2 = 0.00283 x 15.2 / 1000

M2 = 0.000043 M

[OH^-] = 2 x [Sr(OH)2] = 2 x 0.000043 = 0.000086 M

[OH^-] = 0.000086 M

pOH = - log [OH^-]

pOH = - log [0.000086]

pOH = 4.06

pH = 14 - 4.06

pH = 9.94

d) millimoles of OH^- from Sr(OH)2 = 2 (81.0 x 0.000470) = 0.07614

millimoles of OH^- from CsOH = 38.0 x 0.0056 = 0.2128

total millimoles = 0.2128 + 0.07614 = 0.28894

total volume = 81 + 38 = 119 mL

[OH^-] = 0.28894 / 119 = 0.0024 M

[OH^-] = 0.0024 M

pOH = - log [OH^-]

pOH = - log [0.0024]

pOH = 2.62

pH = 14 - 2.62

pH = 11.38