S11-HW2-CH14 - HOMEWORK SET WEEK 2 CHAPTER 14 4.32 a...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
HOMEWORK SET WEEK 2 CHAPTER 14 4.32 a. Soluble; The ions present would be NH 4 + and SO 4 2- (in 2:1 ratio) b. Insoluble c. Insoluble d. Soluble; The ions present would be Ca 2+ and NO 3 - (in 1:2 ratio) 4.34c. Total Ionic equation: Pb 2+ (aq) + 2 NO 3 - (aq) + 2 Na + (aq) + 2 Br - (aq) PbBr 2 (s) + 2 Na + (aq) + 2 NO 3 - (aq) Net Ionic equation: Pb 2+ (aq) + 2 Br - (aq) PbBr 2 (s) This is actually an equilibrium, i.e.: Pb 2+ (aq) + 2 Br - (aq) PbBr 2 (s) ; This type of equilibrium will be discussed further in Expt. #7 and in lecture near the end of term. 14.4 The addition of reactions 1 and 2 yields reaction 3: (Reaction 1) HCN + OH - R R CN - + H 2 O (Reaction 2) H 2 O R R H + + OH - Sum: HCN + OH - + H 2 O CN - + H 2 O + H + + OH - We cancel OH - and H 2 O, which appear on both sides of the equation, to yield: (Reaction 3) HCN R R H + + CN - The rule states that, if a given equation can be obtained from the sum of other equations, the equilibrium constant for the given equation equals the product of the other equilibrium constants. Thus, K for reaction 3 is K = K 1 x K 2 = (4.9 x 10 4 ) x (1.0 x 10 -14 ) = 4.9 x 10 -10 Note that K is small (<< 1), as expected for the ionization of a weak acid. 010. a0. To study the equilibrium, create a saturated solution. Dissolve enough Mg(OH) 2 so that after mixing well, there is still excess solid that settles to the bottom of the container. As long as there is solid present in the flask, equilibrium is established. This solution could then be filtered and titrated with standardized HCl solution to determine [OH ]. [Once we know about pH, we would see that Mg(OH) 2 is a base, so its equilibrium could be studied by pH measurements.] b0. K sp = [Mg 2+ ][OH ] 2 c0. If more solid Mg(OH) 2 is added to a saturated solution, it will settle to the bottom of the container and not dissolve. Remember that pure solids do not appear in the K c expression (see (b)). Therefore, adding more solid will have no effect on the concentrations of Mg 2+ and OH ions in solution. d0. Yes, If a solution is prepared from solid Mg(OH) 2 , then [Mg 2+ ] = 1/2[OH ]. We know this from the stoichiometric coefficients in the balanced equation. In addition, if we know the values of K sp and [OH - ], we can substitute into the K expression to determine [Mg 2+ ]. Presumably, both methods must give the same value for [Mg 2+ ].
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
14.20 The equilibrium is: H 2 O(l) H 2 O(g) and the K p expression is: K p = p(H 2 O(g)) The important concept is the K p expression: it contains the vapor, but not the liquid. The addition of a pure liquid (i.e. water) does not affect the equilibrium. (The pure liquid does not appear in the equilibrium constant). Thus, the amount of water vapor does not change appreciably. This behavior is analogous to the non-effect of adding solid Mg(OH) 2 in Question 14.18c. (More precisely, the amount of vapor decreases slightly because the liquid takes up more
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/20/2011 for the course CHEM 110 taught by Professor Dr.huston during the Spring '06 term at Pittsburgh.

Page1 / 7

S11-HW2-CH14 - HOMEWORK SET WEEK 2 CHAPTER 14 4.32 a...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online