Question

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



(a) 0.0043 M KOH

[OH-] = _____  M

pH = 11.63

(b) 0.0341 g of CsOH in 540.0 mL of solution

[OH -] = 4.21e-4  M

pH = 10.62

(c) 28.1 mL of 0.00187 M Ca(OH)₂ diluted to 800 mL

[OH -] = _____  M

pH = 10.12

(d) A solution formed by mixing 71.0 mL of 0.000480 M Ca(OH)₂ with 24.0 mL of 1.3 x 10^-3 M KOH

[OH -] = _____  M

pH = ______

Answer 1

a)
Given:
[OH-] = [KOH] = 4.3*10^-3 M

use:
pOH = -log [OH-]
= -log (4.3*10^-3)
= 2.3665


use:
PH = 14 - pOH
= 14 - 2.3665
= 11.63
Answer:
[OH-] = 4.3*10^-3 M
pH = 11.63

b)

Molar mass of CsOH,
MM = 1*MM(Cs) + 1*MM(O) + 1*MM(H)
= 1*132.9 + 1*16.0 + 1*1.008
= 149.908 g/mol


mass(CsOH)= 0.0341 g

use:
number of mol of CsOH,
n = mass of CsOH/molar mass of CsOH
=(3.41*10^-2 g)/(1.499*10^2 g/mol)
= 2.275*10^-4 mol
volume , V = 5.4*10^2 mL
= 0.54 L


use:
Molarity,
M = number of mol / volume in L
= 2.275*10^-4/0.54
= 4.212*10^-4 M

SO,
[OH-] = 4.212*10^-4 M

Given:
[OH-] = 4.212*10^-4 M

use:
pOH = -log [OH-]
= -log (4.212*10^-4)
= 3.3755


use:
PH = 14 - pOH
= 14 - 3.3755
= 10.6245
Answer:
[OH-] = 4.212*10^-4 M
pH = 10.6245

c)
use dilution formula
M1*V1 = M2*V2
1---> is for stock solution
2---> is for diluted solution

Given:
M1 = 0.0187 M
V1 = 28.1 mL
V2 = 800 mL

use:
M1*V1 = M2*V2
M2 = (M1*V1)/V2
M2 = (0.0187*28.1)/800
M2 = 6.57*10^-4 M

So,
[OH-] = 2*[Ca(OH)2]
= 2*6.57*10^-4
= 1.31*10^-3 M

Given:
[OH-] = 1.31*10^-3 M

use:
pOH = -log [OH-]
= -log (1.31*10^-3)
= 2.8827


use:
PH = 14 - pOH
= 14 - 2.8827
= 11.1173
Answer:
[OH-] = 1.31*10^-3 M
pH = 11.117

d)
Concentration of mixture = (n1*C1*V1+ n2*C2*V2) / (V1+V2)
where C1 --> Concentration of 1 component
V1-->volume of 1 component
C2 --> Concentration of other component
V2-->volume of other component
n1 --> number of particle from 1 molecule of 1st component
n2 --> number of particle from 1 molecule of 2nd component

Here:
n1 = 2
n2 = 1

use:
C = (n1*C1*V1+ n2*C2*V2) / (V1+V2)
C = (2*4.80*10^-4*71+1*1.3*10^-3*24)/(71+24)
C = 1.05*10^-3 M

Given:
[OH-] = 1.05*10^-3 M

use:
pOH = -log [OH-]
= -log (1.05*10^-3)
= 2.9788


use:
PH = 14 - pOH
= 14 - 2.9788
= 11.0212

Answer:
[OH-] = 1.05*10^-3 M
pH = 11.021