Apa itu Buffer? HA and A- 1. Suatu perisai atau penyangga 2. Suatu asam lemah dan basa konyugatnya: ADA DALAM PROPORSI YANG HAMPIR SAMA 3. Suatu buffer terbuat dari 2 komponen: HA and A- HA + OH- A- + H2O HA + H+ Tdk ada reaksi A- + H+ HA A- + OH- Tdk ada reaksi

A- HA + A- A- HA HA + A- A- HA Hanya, HA A- HA A- HA Pada keadaan Netral HA + A- A- A- A- HA A- HA HA Tambah OH- HA HA + A- HA HA HA A- Tambah H+ A- A- HA + A- (Total buffer) tidak berubah karena HA dan A- berubah hingga jumlah yang sama tapi pada arah yang berlawanan Hanya, Perbandingan dari perubahan A-/HA Penambahan asam atau basa pada suatu buffer merubah perbandingan asam konjugat dan basa tanpa merubah total

Curva Titrasi Titik setengah reaksi pH Daerah Plat NaOH equivalents HCOOH + OH- HCOO- + H2O Asam Konjugat Basa Konjugat NH3 10 NH4+ Titik setengah reaksi 8 [HCOOH] = [HCOO-] pH pKa Daerah Plat 6 Perubahan terkecil dari pH 4 2 HCOO-- HCOOH 0.5 1.0 Curva Titrasi NaOH equivalents

3 7 2 5 5 Menyelesaikan soal Buffer H+ = Ka [HA] [A-] Variabel tetap Variabel bebas pH = pKa + log [HA] [A-] Tetapan 3 7 2 HA + A- + OH- HA + A- + H2O 5 5 Sebelum 2 Sesudah [A-] [HA] = 1.0 [A-] [HA] = 2.33 [A-] + [HA] = 10 [A-] + [HA] = 10

pH menjadi 0.37 satuan lebih alkalin atau basa Soal Buffer ( lanjutan) pH = pKa + log [A-] [HA] Setelah Netralisasi HA = 3 mM Catatan: HA + A- = 10 mM A- = 7 mM = pKa + log 7 3 pH baru adalah: = pKa + log 2.36 = pKa + 0.37 pH menjadi 0.37 satuan lebih alkalin atau basa

Arti Ka Ka is a dissociation constant for an acid, but much more... [H+][A-] [HA] [H+] = Ka [HA] [A-] Rearranging When [HA] = [A-], the acid is half ionized and…. [H+] = Ka Ka is numerically equal to the proton concentration when the acid is half ionized. Adjusting the proton concentration to equal Ka assures the acid will be half ionized.

Significance of pKa 1. The pH at the point of half ionization pKa is the negative log of Ka, i.e., -Log Ka The numerical value for pKa allows one to determine 1. The pH at the point of half ionization 2. Point of maximum buffering capacity 3. Relative Acid strength 4. Order of proton dissociation from a polyprotic acid An acid with a pKa of 4 is 100 times stronger than one with a pKa of 6 A dissociable group with a pKa of 6 is a 1000 times stronger acid than one with a pKa of 9

Problem: 0.1 M acetate buffer has a pH of 5.0. How much NaOH must be added to raise the pH to 5.5? pKa acetate = 4.7 Solution: Calculate the A-/HA at the START Calculate the A-/HA at the END Change in either HA or A will determine base added

Awal Akhir A- = 0.067 mmol A- = 0.086 mmol HA = 0.033 mmol pH = pKa + log [A-] [HA] Awal Akhir [A-] [HA] log = pH - pKa [A-] [HA] log = pH - pKa = pH - pKa = pH - pKa = 5.0 - 4.7 = 5.5 - 4.7 = 0.3 = 0.8 [A-] [HA] = 6.3/1.0 [A-] [HA] = 2.0/ 1.0 A- = 0.067 mmol A- = 0.086 mmol HA = 0.033 mmol HA = 0.014 mmol 0.019 mmol diserap

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?

0.30 0.00 0.52 x 0.52 + x [HCOO-] [HCOOH] [0.52] [0.30] What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Mixture of weak acid and conjugate base! HCOOH (aq) H+ (aq) + HCOO- (aq) Initial (M) Change (M) Equilibrium (M) 0.30 0.00 0.52 -x +x +x 0.30 - x x 0.52 + x pH = pKa + log [HCOO-] [HCOOH] Common ion effect 0.30 – x  0.30 pH = 3.77 + log [0.52] [0.30] = 4.01 0.52 + x  0.52 HCOOH pKa = 3.77 16.2

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) [NH4+] [OH-] [NH3] Kb = = 1.8 X 10-5 0.36 Initial 0.30 + x + x Change - x End 0.30 - x 0.36 + x x (.36 + x)(x) (.30 – x) 1.8 X 10-5 = 0.36x 0.30 1.8 X 10-5  x = 1.5 X 10-5 pOH = 4.82 pH= 9.18 16.3

final volume = 80.0 mL + 20.0 mL = 100 mL Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? final volume = 80.0 mL + 20.0 mL = 100 mL NH4+ 0.36 M x 0.080 L = 0.029 mol / .1 L = 0.29 M OH- 0.050 x 0.020 L = 0.001 mol / .1 L = 0.01M NH3 0.30 M x 0.080 = 0.024 mol / .1 L = 0.24M start (M) 0.29 0.01 0.24 NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq) end (M) 0.28 0.0 0.25 [H+] [NH3] [NH4+] Ka= = 5.6 X 10-10 [H+] = 6.27 X 10 -10 [H+] 0.25 0.28 pH = 9.20 = 5.6 X 10-10 16.3

NH4+ (aq) H+ (aq) + NH3 (aq) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH4+ (aq) H+ (aq) + NH3 (aq) pH = pKa + log [NH3] [NH4+] pH = 9.25 + log [0.30] [0.36] pKa = 9.25 = 9.17 final volume = 80.0 mL + 20.0 mL = 100 mL start (M) 0.29 0.01 0.24 NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq) end (M) 0.28 0.0 0.25 pH = 9.25 + log [0.25] [0.28] = 9.20 16.3

Chemistry In Action: Maintaining the pH of Blood 16.3