Clearly indicating any assumptions/approximations made, calculate the pH, pOH, [

Question

Clearly indicating any assumptions/approximations made, calculate the pH, pOH, [H^+], and [OH^-] of the following solutions. for full credit, including making tables and showing dissociations for each problem. You may not use the "root key" to figure out roots from your calculator (you CAN use the squareroot key). Use tables in the book if needed. Make sure to include units where appropriate. This sheet must be turned in to receive any credit. Each problem is worth 25 points. You may work with one other person and then submit one quiz with both of your names on it: a) 1.5 M caffeine solution K_b caffeine = 5.3*10^-14 b) 2.9*10^-5 M formic acid solution c) 0.024 M perchloric acid solution d) 4.5*10^-2 M lithium hydroxide solution

Explanation / Answer

a)

caffeine is a base so: pKb = 10.4

Kb = 10^-pKb = 10^-10.4 = 3.981*10^-11

B + H2O <-> HB+ + OH-

Kb = [HB+][OH-]/[B]

3.981*10^-11 = x*x/(1.5-x)

x = OH- = 7.73*10^-6

pOH = -log(7.73*10^-6) = 5.111

pH = 14-pOH = 14-5.111 = 8.89

[H+] = 10^-pH = 10^-8.89 = 1.288*10^-9 M

b)

formic acid has a Ka = 1.8*10^-4

so

HA <-> H+ + A-

Ka = [H+][A-]/[HA]

1.8*!0^-4 = x*x/(2.9*10^-5-x)

x = 2.54*10^-5 M = [H+]

pH = -log(2.54*10^-5) = 4.59

pOH = 14-4.59 = 9.41

[OH-] = 10^-9.41 =3.890*10^-10 M

c)

0.024 HClO4

this is a strong acid so

[H+] = [HClO4] =0.024 M

pH = -log(H) = -log(0.024) = 1.61978

pOH = 14-1.61978 = 12.38

[OH-] = 10^-pOH = 10^-12.38 = 4.168*10^-13 M

d)

lithium hyxdroxide is a strong base

LiOH = Lo+ + OH-

[LiOH] =[OH-] = 4.5*10^-2

pOH = -log(OH) = -log( 4.5*10^-2) = 1.3467

pH = 14-pOH = 14-1.3467 = 12.653

[H+] = 10^-pH = 10^- 12.653 =2.223*10^-13

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