Question

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



(a) 0.0034 M RbOH

[OH-] =_________ M

pH = _________

(b) 0.0872 g of KOH in 510.0 mL of solution

[OH -] =__________ M

pH = __________

(c) 79.8 mL of 0.00719 M Sr(OH)₂ diluted to 900 mL

[OH -] = __________ M

pH = ____________

(d) A solution formed by mixing 64.0 mL of 0.000880 M Sr(OH)₂ with 36.0 mL of 2.6 x 10^-3 M RbOH

[OH -] = ________M

pH =__________

Answer 1

(a)

0.0034 M RbOH

RbOH - rubidium hydroxide is a strong base and hence it will dissociate completely.

so, [OH-] = 0.0034 M

pOH = - log [OH-] = - log(0.0034) = 2.47

pH = 14 - pOH = 14 - 2.47 = 11.53

(b)

0.0872 g of KOH in 510.0 mL of solution

Molar mass of KOH = 56.11 g/mol

So, 56.11 g of KOH = 1 mole

0.0872 g of KOH = (0.0872/56.11) mole = 0.00155 mole

Volume = 510.0 mL = 0.510 L

Molarity = 0.00155 mole / 0.510 L = 0.003 M

KOH - Potassium hydroxide is a strong base and hence it will dissociate completely.

so, [OH-] = 0.003 M

pOH = - log [OH-] = - log(0.003) = 2.52

pH = 14 - pOH = 14 - 2.52 = 11.48

(c)

79.8 mL of 0.00719 M Sr(OH)2 diluted to 900 mL

We know that

M1V1 = M2V2

M2 = M1V1 / V2

= (79.8 mL) (0.00719 M) / (900 mL)

= 0.00064 M

Sr(OH)₂ is a strong base and hence it will dissociate completely.

Sr(OH)₂     Sr⁺² + 2OH^-

so, [OH-] = 2 x 0.00064 M = 0.00128 M

pOH = - log [OH-] = - log(0.00128) = 2.89

pH = 14 - pOH = 14 - 2.89 = 11.11

(d)

64.0 mL of 0.000880 M Sr(OH)2

Moles of Sr(OH)2 = 0.000880 M x 0.064 L = 0.00005632 moles

1 mole of Sr(OH)2 contains 2 moles of OH^- ion.

0.00005632 moles of Sr(OH)2 contains 2(0.00005632) moles of OH^- ion.

0.00005632 moles of Sr(OH)2 contains 0.00011264 moles of OH^- ion.

36.0 mL of 2.6 x 10^-3 M RbOH

Moles of RbOH = 2.6 x 10^-3 M x 0.036 L = 0.0000936 moles

1 mole of RbOH contains 1 mole of OH^- ion.

0.0000936 moles of RbOH contains 0.0000936 mole of OH^- ion.

Total moles of OH^- ion = 0.00011264 mole + 0.0000936 mole

= 0.00020624 mole

Total volume = 64.0 mL + 36.0 mL = 100 mL = 0.1 L

[OH^-] = 0.00020624 mole / 0.1 L = 0.0020624 M

pOH = - log [OH-] = - log(0.00020624) = 3.69

pH = 14 - pOH = 14 - 3.69 = 10.31