Question

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



(a) 0.0013 M NaOH

[OH-] = ____M

pH =_____

(b) 0.0571 g of CsOH in 540.0 mL of solution

[OH -] ____= M

pH =____

(c) 11.6 mL of 0.00247 M Ba(OH)₂ diluted to 800 mL

[OH -] = ___ M

pH =___

(d) A solution formed by mixing 82.0 mL of 0.000500 M Ba(OH)₂ with 54.0 mL of 6.4 x 10^-3 M NaOH

[OH -] = ___M

pH =____

Answer 1

We will apply below formulas,

[H+] * [OH-] = 10^-14

pH + pOH = 14

pH = - log[H+]

pOH = - log [OH-]

Since, all bases are strong and hence [Base] = [OH-]

A) 0.0013M NaOH

[OH-] = 0.0013 M

pOH = - log (0.0013) = 2.886

pH = 14 - 2.886 = 11.114 ....Answer

B) Moles of CsOH = mass/molar mass = 0.0571/149.913 = 0.000381 moles

[CsOH] = moles/volume in liters = 0.000381/0.540 = 0.000705 M

[CsOH] = [OH-] = 0.000705 M

pOH = - log (0.000705) = 3.152

pH = 14 - 3.152 = 10.85 ....Answer

C) Moles of Ba(OH)2 = Molarity * volume in liters = 0.00247 * 11.6/1000 = 0.00002865

New molarity i.e. [Ba(OH)2] = moles/volume in liters = 0.00002865/0.800 = 0.00003582 M

Since, Ba(OH)2 contains 2 OH-

Thus, [OH] = 2 * 0.00003582 = 0.00007163 M

pOH = - log (0.00007163) = 4.145

pH = 14 - 4.145 = 9.85 ...Answer

Let me know if any doubts.

Please note that as per HOMEWORKLIB POLICY, I have answered first 3 parts.