Question

Calculate [OH ^-] and pH for each of the following solutions.



(a) 0.0062 M LiOH

[OH-] = M

pH =

(b) 0.0431 g of LiOH in 480.0 mL of solution

[OH -] = M

pH =

(c) 13.7 mL of 0.00220 M Sr(OH)₂ diluted to 600 mL

[OH -] = M

pH =

(d) A solution formed by mixing 15.0 mL of 0.000560 M Sr(OH)₂ with 89.0 mL of 1.2 x 10^-3 M LiOH

[OH -] = M

pH =

Answer 1

(a)

0.0062 M LiOH

LiOH ------->   Li+ + OH-

Thus, [OH-] = 0.0062 M

pOH = -log [OH-]

pOH = 2.21

pH = 14 - pH = 14 - 2.2 = 11.8

Thus, [OH-] = 0.0062 M and pH = 11.8

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(b)

0.04321 g of LiOH in 480.0 mL of solution.

Let us find the concentration of LiOH from the given information.

Molarity = [Moles / Volume] x 100

Moles = Mass / Molar mass = 0.04321 g / 23.9 g/mol = 1.81 x 10^-3 mol

Molarity = (1.81 x 10^-3 mol) / 0.480 = 3.77 x 10^-3 M

LiOH ------->   Li+ + OH-

Thus, [OH-] = 3.77 x 10^-3 M

pOH = -log [OH-]

pOH = 2.4

pH = 14 - pH = 14 - 2.4 = 11.6

Thus, [OH-] = 3.77 x 10^-3 M and pH = 11.6

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c)

13.7 mL of 0.00220 M Sr(OH)2 diluted to 600 mL

We use the equation M1V1 = M2V2, inorder to find the molarity of solution after dilution.

So, (0.00220 M) ( 13.7 mL) = (M2) ( 600mL)

M2 = 5.02 x 10^-5 M

Sr(OH)2 ------------> Sr^2+ + 2OH-

Thus, [OH-] = 2 x 5.02 x 10^-5 = 1 x 10^-4 M

pOH = -log [OH-]

pOH = 3.9

pH = 14 - pOH = 14 - 3.9 = 10.1

Thus, [OH-] = 1 x 10^-4 M and pH = 10.1

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(d)

15.0 mL of 0.000560 M Sr(OH)2 with 89.0 mL of 1.2 x 10^-3 M LiOH

Moles of Sr(OH)2 = Molarity x Volume(in L ) = 0.000560 x 0.015 = 8.4 x 10^-6 mol

Sr(OH)2   -------------> Sr^2+ + 2OH-

Thus, moles of OH- = 2 x 8.4 x 10^-6 = 1.68 x 10^-5 mol

Moles of LiOH = Molarity x Volume(in L ) = 1.2 x 10^-3 x 0.089 = 1.1 x 10^-4 mol

LiOH -----------> Li+ + OH-

Thus, moles of OH- = 1.152 x 10^-4 mol

[OH-] = Total moles of OH- / Total volume(in L) = 1.152 x 10^-4 / 0.104 = 1.11 x 10^-3 M

pOH = -log [OH-]

pOH = 2.95

pH = 14 - pOH = 14 - 2.95 = 11.05

Thus, [OH-] = 1.11 x 10^-3 M and pH = 11.05