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 270.0mL of pure water, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

Part B- For 270.0mL of a buffer solution that is 0.215M in HCHO2 and 0.285M in KCHO2, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

Part C- For 270.0mL of a buffer solution that is 0.305M in CH3CH2NH2 and 0.275M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

Explanation / Answer

a)
pure water : pH = 7.00 (initial pH)
moles NaOH = 0.010
[OH-]= 0.010 mol/ 0.270 L=0.037 M
pOH = - log 0.037 = 1.43
pH = 14 - pOH = 14 - 1.43
pH = 12.57 (final pH)

(b)
Ka = 1.8 x 10-4
pKa = 3.74
pH = 3.74 + log 0.285 / 0.215 = 3.86 (initial pH of the buffer)

moles formic acid = 0.215 M x 0.270 L = 0.058
moles formate = 0.315 M x 0.270 L =0.07695
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 = 0.058 - 0.010 = 0.048
moles HCOO- = 0.07695 + 0.010 = 0.08695
concentration HCOOH = 0.048 / 0.270 L=0.178 M
concentration HCOO- = 0.08695 / 0.270 = 0.322
pH = 3.74 + log 0.322 / 0.178 = 4.0 ( final pH)

(c)
Kb of ethylammine = 4.3 x 10-4
pKb = 3.37
pOH = 3.37 + log 0.275 / 0.305 = 3.325
pH = 14 - 3.325 = 10.675 ( initial pH)
moles CH3CH2NH3+ = 0.275 x 0.270 L = 0.07425
moles CH3CH2NH2 = 0.305 x 0.270 L = 0.08235
CH3CH2NH3+ + OH- = CH3CH2NH2 + H2O
moles CH3CH2NH3+ = 0.07425 - 0.010 =0.06425
moles CH3CH2NH2 = 0.08235 + 0.010 =0.09235
concentration CH3CH2NH3+ = 0.06425 / 0.270 =0.238 M
concentration CH3CH2NH2 = 0.09235 / 0.270 = 0.342 M
pOH = 3.37 + log 0.238 / 0.342 = 3.21
pH = 14 - 3.21 = 10.79 (final pH)

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