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Lecture 31 Acids and Bases III Tutorial

Lecture 31 Acids and Bases III Tutorial

www.apchemsolutions.com Lecture 31 Acids and Bases III Tutorial

Write balanced net ionic equations for the reactions that occur in questions 1-5.

1) Equal volumes of 0.15 M and 0.1.5 M lithium are mixed.

- - H2SO3(aq) + OH (aq) Æ HSO3 (aq) + H2O(l)

2) Solutions of ammonium nitrate and sodium cyanide are poured into a beaker.

+ - NH4 (aq) + CN (aq) Æ NH3(aq) + HCN(aq)

3) Carbon dioxide gas is bubbled through a solution of 0.5 M .

- - CO2(g) + OH (aq) Æ HCO3 (aq)

4) bromide gas is mixed with ammonia gas.

HBr(g) + NH3(g) Æ NH4Br(s)

5) Solid sodium acetate is stirred into a beaker of distilled water.

+ - NaCH3CO2(s) + H2O(l) Æ Na (aq) + CH3COOH(aq) + OH (aq)

6) Which mixture, 1.5 M NH4Cl or 1.5 M NaCl, has the lowest pH? Justify your answer.

+ 1.5 M NH4Cl will have the lowest pH. Its pH will be less than 7. NH4 is an acidic anion. Acidic anions lower the pH.

Cl- is a neutral anion, which will not affect the pH.

1.5 M NaCl will have the highest pH. NaCl and water produces a neutral solution with a pH of 7, as both ions are neutral. Na+ is the conjugate acid of a strong base, and Cl- is the conjugate base of a strong acid.

© 2009, 2008 AP Chem Solutions. All rights reserved. 1 www.apchemsolutions.com 7) One mixes 150 mL of 0.3 M HCl with 150 mL of 0.4 M LiOH. What is the pH of the solution?

[OH-] = 0.3 – 0.4 = 0.1 M

pOH = -log[OH− ] =− log(0.1) = 1

pH = 14 - pOH = 14 - 1 = 13

8) The acid dissociation constants for , HOI, and lactic acid, -11 -4 HC3H5O3, at 298 K are 2 x 10 and 1.38 x 10 respectively. Which solution is more basic: 1.0 M NaOI or 1.0 M NaC3H5O3? Justify your answer.

HOI is the weaker acid, as 2 x 10-11 < 1.38 x 10-4. This means that its conjugate base, OI- - , is stronger than C3H5O3 . Weaker acids have stronger conjugate bases.

- - Because OI is a stronger base than C3H5O3 , the equilibrium for - - OI (aq) + H2O(l) U HOI(aq) + OH (aq) - - lies further to the right than it does in C3H5O3 (aq) + H2O(l) U HC3H5O3(aq) + OH (aq).

Thus, 1.0 M NaOI produces a higher concentration of OH-, making a more basic solution.

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