For each of the following solutions, calculate the initial pH and the final pH a

Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.010 mol of NaOH . Part A For 230.0 mL of pure water, calculate the initial pH and the final pH after adding 0.010 mol of NaOH . Express your answers using two decimal places separated by a comma. p H initial , p H final = SubmitRequest Answer Part B For 230.0 mL of a buffer solution that is 0.215 M in HCHO 2 and 0.315 M in KCHO 2 , calculate the initial pH and the final pH after adding 0.010 mol of NaOH . Express your answers using two decimal places separated by a comma. p H initial , p H final = SubmitRequest Answer Part C For 230.0 mL of a buffer solution that is 0.280 M in CH 3 CH 2 NH 2 and 0.250 M in CH 3 CH 2 NH 3 Cl , calculate the initial pH and the final pH after adding 0.010 mol of NaOH . Express your answers using two decimal places separated by a comma.

Explanation / Answer

a)
pure water : pH = 7.00 (initial pH)
moles NaOH = 0.010
[OH-]= 0.010 mol/ 0.230 L=0.0435 M
pOH = - log 0.0435 = 1.36
pH = 14 - pOH = 14 - 1.36
pH = 12.64 (final pH)

(b)
Ka = 1.8 x 10-4
pKa = 3.74
pH = 3.74 + log 0.315 / 0.215 = 3.906 (initial pH of the buffer)

moles formic acid = 0.215 M x 0.230 L = 0.0495
moles formate = 0.315 M x 0.230 L =0.0725
the effect of the added 0.010 mol OH- would be to decrease the moles of formic acid by 0.010 and increases the moles of formate by 0.010 by the reaction
HCOOH + OH- = HCOO-
moles HCOOH =0.0495 - 0.010 = 0.0395
moles HCOO- = 0.0725 + 0.010 = 0.0825
concentration HCOOH = 0.048 / 0.230 L=0.172 M
concentration HCOO- = 0.0825 / 0.270 = 0.3586 M
pH = 3.74 + log 0.3586 / 0.172 = 4.06 ( final pH)

(c)
Kb of ethylammine = 4.3 x 10-4
pKb = 3.37
pOH = 3.37 + log 0.25 / 0.28 = 3.321
pH = 14 - 3.321 = 10.679 ( initial pH)
moles CH3CH2NH3+ = 0.25 x 0.230 L = 0.0575
moles CH3CH2NH2 = 0.28 x 0.230 L = 0.0644
CH3CH2NH3+ + OH- = CH3CH2NH2 + H2O
moles CH3CH2NH3+ = 0.0575 - 0.010 =0.0475
moles CH3CH2NH2 = 0.0644 + 0.010 =0.0744
concentration CH3CH2NH3+ =0.0475 / 0.230 =0.2065 M
concentration CH3CH2NH2 = 0.00.0744 / 0.230 = 0.323 M
pOH = 3.37 + log 0.2065 / 0.323= 3.176
pH = 14 - 3.21 = 10.82 (final pH)

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