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Acids and Bases

Acids and Bases. Matt Marcin Jason Reinglass. Acids. Any substance that when dissolved in pure water, increases the concentration of hydrogen ions, H + (aq) React with bases in aqueous solution to produce a salt and water.

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Acids and Bases

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  1. Acids and Bases Matt Marcin Jason Reinglass

  2. Acids • Any substance that when dissolved in pure water, increases the concentration of hydrogen ions, H+(aq) • React with bases in aqueous solution to produce a salt and water. • A substance that contains hydrogen and releases a hydrogen ion (H+) as one of the products of ionic dissociation in water. • Stronger the acid, the weaker the conjugate base. • Polyprotic acids are capable of donating two more or protons. • Strong Acids ionize (dissociate) 100% in aqueous solutions • HCl, H2SO4, HBr, HI, HNO3, and HClO4.

  3. Bases • Any substance that increases the concentration of hydroxide ion, OH-(aq) when dissolved in pure water. • Have bitter taste, cause metal ions to form insoluble compounds that precipitate from solution. • Polyprotic bases are capable of accepting more than one proton. • Strong Bases: • NaOH, C2H5O-, NH2-, H-, CH3-

  4. Titrations • Equivalence Point- point where number of moles of OH- added exactly equals the number of moles of H+ that can be supplied by the acid. • When the equivalence point has been reached in a titration, the volume of base added since the beginning of the titration can be determined by reading the calibrated buret. • Standardization- procedure by which the concentration of an analytical reagent is determined accurately. • #1. Weigh accurately a sample of a pure, solid acid or base, and then titrate this sample with a solution of the base or acid to be standardized. • #2. To standardize a solution is to titrate it with another solution that is already standardized.

  5. Titration Example • The entire contents of a 354 mL can of Diet Dr. Thunder (which has been allowed to “go flat” so it isn’t carbonated) is titrated w/ 0.621 M calcium hydroxide solution; 27.34 mL of the titrant is needed to completely neutralize the acid in the drink. Assume the drink is acidic entirely due to dissolved phosphoric acid, H2PO4, which is triprotic and has Ka1= 7.5 x 10-3.

  6. Titration Example Cont’ • Write a balanced reaction equation for the reaction between phosphoric acid and calcium hydroxide. 2H3PO4 + 3 Ca(OH)2 → Ca3(PO4)2 + 6 H2O

  7. Titration Example Cont’ • Calculate the phosphoric acid concentration in the drink. nAVACA = nBVBCB 3(354mL)CA = 2(27.34mL)(0.621M) CA = 0.03197 M (≈0.0320 M)

  8. Titration Example Cont’ • Calculate the pH of the original (untitrated) drink. H3PO4 + H2O → H3O+ + H2PO4- 0.03197 x x Ka = 7.5 x 10-3 = X2/0.03197 => X= √(0.03197)(7.5 x 10-3) = 0.01549M = [H+] pH= -log [H+] = 1.81

  9. Bronsted Acids and Bases/Values of K • Amphiprotic Substance- molecules or ions that can behave as either a Bronsted Acid or base (ex. Water) • Bronsted Acid-Base Theory: • All proton transfer reactions proceed from the strongest acid-base pair to the weakest acid base pair. • Ka=[H3O+][A-]/[HA] • Kb=[BH+][OH-]/[B] • The value of K is less than one for a weak acid • Value of K becomes smaller and smaller with each successive ionization step. • Large value of K indicates products are strongly favored, whereas a small k indicates reactants are favored. • Kw=[H3O+][OH-] Kw=1*10^-14

  10. Other Acid/Base Concepts • Hydrolysis-conjugate base steals proton from H2O so therefore conjugate base gains an H+ • Buffer-function of it is to resist changes in pH. • Depending on what is added, pH barely changes/effected. • Create a stable system which doesn’t easily change pH. • Common Ion Effect-when an ion “common” to the ionization of the acid is present in an amount greater than that produced by simple acid ionization. • More difficult to remove H+ from negatively charged ion such as H2PO4- then from a neutral molecule such as H3PO4. • Reactions always proceed in direction of weaker acid-base pair.

  11. Buffer Example • A buffer is prepared by dissolving 16.50 g of sodium benzoate in 500. mL of 0.250 M benzoic acid (C6H5COOH). Benzoic acid’s Ka is 6.3 x 10-5. Assume the volume change is negligible.

  12. Buffer Example Cont’ • Calculate pH of prepared solution. pKa = 4.201 pH= pKa + log [C6H5COO-]/ [C6H5COOH] = 4.201 + log [.230 M]/ [.250 M] pH = 4.16 (.230 M sodium benzoate was found by converting 16.50 g to moles and then dividing by .500 L)

  13. Buffer Example Cont’ • Calculate the pH after 2.50 g of solid NaOH has been added to 500. mL of buffer solution. (convert 2.50 g NaOH to moles and divide by .500 L) .125 M (500mL) = 62.505 mmoles .250 M (500mL)= 125 mmoles .230 M (500mL)= 115 mmoles pH= pKa + log [C6H5COO-]/ [C6H5COOH] pH=4.201 + log [(115+62.505)500mL]/ [(125-62.505)500mL] pH=4.201 + .453 = 4.65

  14. Buffer Example Cont’ • Calculate how much 1.00 M HCl solution would need to be added to 500. mL of the original buffer solution to lower its pH by 1.00 pH units. pH= pKa + log [C6H5COO-]/ [C6H5COOH] 4.16-1=4.201 + log [115-X]/ [125+X] -1.041 = log [115-X]/[125+X] 10-1.041=115-X/125+X 0.09099=115-X/125+X .09099(125+X) = 115-X X= 94.4835 mmoles HCl 94.4835 mmoles HCl (1L/1 mole HCl) 94.5 mL of HCl solution

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