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#25 Key #25 Key Document Transcript

  • KEY GENERAL CHEMISTRY-II (1412) S.I. # 25 1. Define the common ion effect: the extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte. 2. Define buffer capacity the buffer capacity is how much acid or base can neutralize before the pH begins to change too much. 3. Define a buffer solution is a solution of (1) a Weak acid or a Weak base and (2) its salt both components must be present. •A buffer solution has the ability to resist changes in pH 4. Define the pH range the pH depends on Ka for the acid and the relative concentration of acid and base that compromise that buffer. The pH range of any buffer is the ph range over which the buffer acts effectively. 5. Provide the Henderson-Hasselbalch Equation pH = pKa + log [base] [acid] used to calculate pH in Buffer solutions 6. What is the pH of a buffer that is 0.12M in lactic acid (HC3H5O3) and 0.10 M in sodium lactate? For lactic acid, Ka = 1.4 x 10-4. Conjugate base of HC3H5O3 is the lactate ion C3H5O3- Get the pH using the ICE method: HC3H5O3 (aq)  H+ (aq) + C3H5O3- (aq) Initial 0.12M 0 0.10 Change -x M +xM +xM Equilibrium (0.12 – x) M xM (0.10 – x) M Ka = 1.4x10-4 = [H+][ C3H5O3] = x(0.10 + x) [C3H5O3] (0.12 – x) Ka is small and a common ion is present, we expect x to be small relative to either 0.12 or 0.1 M so our equation can be simplified to give Ka = 1.4x10-4 = x(.10) 0.12 [H ] = x = 0.12 (1.4x10 ) = 1.7x10-4 + -4  pH = -log(1.7x10-4) = 3.77 0.1 Henderson Hasselbalch equation could be used to calculate pH directly pH = pKa + log([base]/[acid]) = 3.85 + log(0.1/0.12) = 3.77
  • KEY 7. Calculate the molar concentration of OH- ions in a 1.15 M solution of hypobromite ion (BrO-); Kb = 4.0 x 10-6 . What is the pH of this solution? BrO- (aq) + H2O(l)  HOBr (aq) + OH- (aq) Initial 1.15M 0 0 Equil. (1.15 – x)M xM xM Kb = [HOBr][OH-] = x2 ~ x2 = 4.0x10-6 - [BrO ] 1.15-x 1.15 x2 = 1.15(4.0x10-6); x=[OH-] = 2.14x10-3 = 2.1x10-3 M; pH=11.33 8. Calculate the pH of a buffer that is 0.120 M in NaHCO3 and 0.105 M Na2CO3. The conjugate acid in this buffer is HCO3- so use Ka2 for H2CO3, 5.6x10-11 HCO3- (aq)  H+ (aq) + CO3-2 (aq) Ka = [H+ ][ CO3-2] so [H+] = Ka [HCO3-] = 5.6x10-11 (0.120) [HCO3-] [ CO3-2] (0.105) [H+] = 6.40 x 10 -11 = pH = 10.19 9. Explain the difference between solubility and solubility-product constant. Write the expression for ionic compounds: MnCO3, Hg(OH)2 and Cu3(PO4)2. Solubility is the amount (grams, moles) of solute that will dissolve in a certain volume of solution. Solubility-product constant is an equilibrium constant, the product of the molar concentrations of all the dissolved ions in solution See sample exercise 17.9 Ksp = [Mn2+][CO32-]; Ksp = [Hg+][OH-]2 Ksp = [Cu2+]3[PO43-]2 10. Which of the following salts will be substantially more soluble in acidic solution than in pure water? A. ZnCO3 B. ZnS C. BiI3 D. AgCn E. Ba3(PO4)2 If the anion of the salt is the conjugate bae of a weak acid, it will combine with H+ reducing the concentration of the free anion in solution, causing more salt to dissolve. More soluble in acid: All but C.