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

Question

For each of the following solutions, calculate the initialand the finalafter adding 0.010 mol of NaOH.

A)

For 210.0mL of pure water, calculate the pH initial and the pH final after adding 0.010 mol of NaOH.

B)

For 210.0 mLof a buffer solution that is 0.215M in HCHO2 and 0.280M in KCHO2, calculate the pH initial and the pH final after adding 0.010 mol of NaOH.

C)

For 210.0mL of a buffer solution that is 0.325M in CH3CH2NH2 and 0.295M in CH3CH2NH3Cl, calculate the pH initial and the pH final after adding 0.010 mol of.

Explanation / Answer

A. Pure water should have pH of 7(initial pH).
Final pH: 0.010 mol NaOH/ 0.300 L = 0.0333 M NaOH
It is a strong base, so it fully dissociates into OH- and Na+ ions
NaOH <--> Na+ + OH-
[NaOH] = [OH-]
-log[OH-] = pOH
-log[0.0333] = 1.477
14 - 1.477 = pH = 12.5

B. Moles HCHO2: 0.245 mol/L x 0.240 L = 0.0588 moles weak acid
Moles KCHO2: 0.310 mol/L x 0.240 L = 0.0744 moles conj. base

Strong base NaOH will convert some HCHO2 into conjugate base.
Moles NaOH = moles CHO2-
Moles HCHO2 left: 0.0588 mol - 0.010 mol = 0.0488 mol
Moles conj. base total: 0.0744 + 0.010 mol = 0.0844 mol

Henderson-Hasselbalch equation:
pH = pKa + log([KCHO2]/[HCHO2])
initial pH = 3.745 + log(0.310/0.245)
initial pH = 3.85

Final pH = 3.745 + log(0.0844/0.0488)
Final pH = 3.98

C. Moles CH3CH2NH2: 0.260 M x 0.300 L = 0.078 mol
Moles conjugate acid: 0.235 M x 0.300 L = 0.0705 mol

pKa acid I'm using: 10.63
Initial pH = 10.63 + log(0.078/0.0705)
Initial pH = 10.7

Adding 0.010 mol NaOH will convert some CH3CH2NH3+ into
conjugate base, CH3CH2NH2.
Total moles CH3CH2NH2: 0.010 mol + 0.078 mol = 0.088 mol
Moles acid left: 0.0705 mol - 0.010 mol = 0.0605 mol

Final pH = 10.63 + log(0.088/0.0605)
Final pH = 10.8

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