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Strong base neutralizes weak acid

Strong base neutralizes weak acid. Strong acid neutralizes weak base. How do we calculate the pH of a buffer ? 2 ingredient problem 0.10 M CH 3 COOH + 0.20 M CH 3 COO - pH = 5.05. NaA ( s ) Na + ( aq ) + A - ( aq ). HA ( aq ) H + ( aq ) + A - ( aq ). K a [HA].

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Strong base neutralizes weak acid

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  1. Strong base neutralizes weak acid Strong acid neutralizes weak base

  2. How do we calculate the pH of a buffer ? 2 ingredient problem 0.10 M CH3COOH + 0.20 M CH3COO- pH = 5.05

  3. NaA (s) Na+(aq) + A-(aq) HA (aq) H+(aq) + A-(aq) Ka [HA] [H+][A-] Ka = [H+] = [HA] [A-] -log [H+] = -log Ka - log [HA] [A-] -log [H+] = -log Ka + log [HA] [A-] [conjugate base] pH = pKa + log [acid] Consider mixture of salt NaA and weak acid HA. Henderson-Hasselbalch equation Take log Assumption: Changes in [A-] & [HA] will be negligible (within 5%) if Ka < 0.01 & Ka < 0.01 [base] [acid] use initial conc. of acid and base to calculate pH Remember pX = -logX so pKa = -log Ka [A-] pH = pKa + log [HA]

  4. HCOOH (aq) H+(aq) + HCOO-(aq) Initial (M) Change (M) Equilibrium (M) [HCOO-] pH = pKa + log [HCOOH] [0.52] pH = 3.77 + log [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! 0.30 0.00 0.52 -x +x +x 0.30 - x x 0.52 + x Common ion effect 0.30 – x 0.30 = 4.01 0.52 + x 0.52 HCOOH pKa = 3.77 16.2

  5. Consider an equal molar mixture of CH3COOH and CH3COONa H+(aq) + CH3COO-(aq) CH3COOH (aq) OH-(aq) + CH3COOH (aq) CH3COO-(aq) + H2O (l) • A buffer solution is a solution of: • A weak acid or a weak base and • The salt of the weak acid or weak base • Both must be present! A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. Add strong acid Add strong base 16.3

  6. Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3 (a) KF is a weak acid and F- is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO32- is a weak base and HCO3- is it conjugate acid buffer solution 16.3

  7. NH4+(aq) H+(aq) + NH3(aq) pH = 9.25 + log [0.30] [NH3] pH = pKa + log [NH4+] [0.36] 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? pKa= 9.25 = 9.17 Ka= 5.6 x 10-10 => pOH = 14 - pH = 4.75 [OH-] = 1.8 x 10-5 M

  8. [0.25] 0.025 0.028 [0.28] 0.10 0.10 NH4+(aq) + OH-(aq) H2O (l) + NH3(aq) pH = 9.25 + log [NH3] = [NH4+] = Calculate moles of all components. Remember we have 80.0 mL of the buffer solution @ 0.30 M NH3/0.36 M NH4Cl and 20.0 mL of 0.050 M NaOH. Moles NH4+= (0.080 L)(0.36 M) = 0.029 moles Moles NH3 = (0.080 L)(0.30 M) = 0.024 moles Moles NaOH = (0.020 L)(0.050 M) = 0.0010 moles final volume = 80.0 mL + 20.0 mL = 100 mL 0.024 start (moles) 0.029 0.0010 end (moles) 0.028 0.0 0.025 Moles/Volume(L) = 9.20 ΔpH = 0.03

  9. Maintaining the pH of Blood 16.3

  10. How do we build a better buffer? Add approximately equal quantities of acid and base Have relatively high concentrations of acid and base => the larger the [acid] & [base] the greater the buffer capacity How do we prepare a buffer at a given pH? Choose acid/base conjugate pair from table Check to be sure that they are unreactive in the system used pKa pH typical rule of thumb pH + 1 = pKa

  11. [conjugate base] pH = pKa + log [acid] Create buffer with pH = 7.50. pH + 1 = pKa Look for pKa‘s 6.5 => 8.5 => Ka‘s 3 x 10-7 => 3 x 10-7 HOCl / OCl Ka = 3.5 x 10-8 H2PO4- / HPO4-2Ka2 = 6.2 x 10-8 H2AsO4- / HAsO4-2Ka2 = 8 x 10-8 H2CO3 / HCO3-Ka1 = 4.3 x 10-7 Calculate quantities of acid and base.

  12. [HCO3-] [H2CO3] H2CO3 / HCO3-Ka1 = 4.3 x 10-7 pKa1 = 6.37 pH - pKa = log 7.50 - 6.37 = 1.13 [HCO3-] 10-1.13 = 13.6 = [H2CO3] Set [H2CO3] = 0.0100 M (NOTE: This is a judgement call.) then [HCO3-] = 13.6 [H2CO3] = 0.136 M

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