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 =____ |
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.
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