Chemistry questions (various)

Question

I'm reviewing for an exam, and i'm wondering if any of you could answer this questions for me. What i need to see is HOW you got these answers, as i already have an answer sheet, just not the solutions.

1. The pH of 0.050M aqueous solution of calcium hydroxide

2. For the reaction C2H6-> 2CH3(g) rate=k[C2H6]

if K= 5.50*10^-4 and [C2H6]intial=0.0200M calculate the rate reaction after 1 hour.

3. given 4PH3(g)-> P4(g) 6H2(g) rate=k[PH3]

if the rate constant is 0.0278s-1, calculate the percent of the orginal PH3 which reacted after 76.0 s

Remember, please show as much work as you can, so i can solve these problems on my own. Thanks!

Answer 1

1. The pH of 0.050M aqueous solution of calcium hydroxide

Ca(OH)2 --> Ca+2 & 2OH-

Ksp = [Ca+2] [OH-]^2

6.5e-6 = [0.05] [OH-]^2

OH- = 0.0114

pOH = 1.94

pH = 14 - pOH = your answer : 12.06

======================================...

2. For the reaction C2H6-> 2CH3(g) rate=k[C2H6]

if K= 5.50*10^-4 and [C2H6]intial=0.0200M calculate the rate reaction after 1 hour.

for first order reactions:

log [At] / [Ao] = - kt/ 2.303

log [At] / [Ao] = - (5.50E-4)(3600 seconds)/ 2.303

log [At] / [Ao] = - 0.860

1O^x both sides

[At] / [Ao] = 0.138

[At] = 0.138 [Ao]

[At] = 0.138 [0.0200] = 2.76e-3 Molar after 1 hour

----------------------

rate=k[PH3]

rate= 5.50*10^-4 [2.76e-3]

your answer: rate = 1.52 e-6 mol/litre-sec

======================================...

3. given 4PH3(g)-> P4(g) 6H2(g) rate=k[PH3]

if the rate constant is 0.0278s-1, calculate the percent of the orginal PH3 which reacted after 76.0 s

log [At] / [Ao] = - kt/ 2.303

log [At] / [Ao] = - (0.0278s-1)(76 seconds)/ 2.303

log [At] / [Ao] = - 0.917

1O^x both sides

[At] / [Ao] = 0.1209

% = [At] / [Ao] times 100 = 12.09 % is left

your answer(3sigfigs): 12.1%

Answer 2

1. the equation to find the ph is -log[H+ ions], so if you plug in the Molarity into the equation for the H+ ions, you would do -log[.050] which equals 1.301.

I'm sorry that's the only one I know how to do, Good Luck Though!