To make a long story short, my youngest daughter decided to endure 
through 4 semesters of chemistry, and now is paying up for it. She is 
a talented little girl, but science is not her thing and it hurts for 
me to see her study so hard but understand so little. It runs in the 
family. I wanted to help myself, but me and my wife are no scientists 
at all, plus it was so long ago. Tutoring is very very weak at where 
she goes, and we know few friends around here who are capable of 
helping, most of which already are looking at us the wrong way for 
asking so much. I decided to open my pocket and let you, Chemistry 
folks, help us out. I secretly took the next set of questions from my 
daughters assignment section, and here they are. I understand that 
many (or most) of you would have the problem to giving direct help 
like this, "learning and understanding the material is the most 
important", right ? Wrong. Only but a few of you would understand the 
pain I feel when I see my daughter quietly cry in her room over her 
low assignment grades after hours and hours of studying daily. 
  
Below I copied down some of the questions and choice answers that came 
along. I tried my best to copy down everything number to number, word 
for word. Its a little bit difficult to transfer everything to text, 
but I tried. Please feel free to clarify if you did not understand a 
part. 
  
I am asking for a list of answers to the problems, in numerical + 
content format, as such : 
  
Question 1: Answer 3. The number of moles is 4.0.  
Question 2: Answer 5. The reaction is reversed.  
  
Time is an issue, but I would like the answers to be correct. I do not 
want to check my daughters answers and tell her that one of her 
correct answers is incorrect simply because what was given to me is a 
wrong answer. Once again I apologize for such rude question, but I am 
desperate, and I ran out of all other options. I am also considering a 
big tip for those who'll work hard on these, do all (or most) of the 
questions, provide the correct answers and hopefully as soon as 
possible. 
  
Thank you once more. You are my last hope. 

1) What is the pH of a 0.020 M solution of 
hydrosulfuric acid, a diprotic acid?

Ka1 = 1.1 * 10^-7 Ka2 = 1.0 * 10^-14

1. 3.65
2. 4.33
3. 4.69
4. 5.22
5. 7.00
6. 7.84
7. 9.67

2) What is the H3O+ concentration of a 0.15 M
solution of NH4Cl in H2O at 25deg Celcius ? (Kb for
NH3 = 1.8 * 10^-5)

1. 1.0 * 10^-14 M
2. 1.1 * 10^-9 M
3. 1.6 * 10^-3 M
4. 4.9 * 10^-5 M
5. 8.2 * 10^-1 M
6. 9.1 * 10^-6 M

3) The pH of a 0.10 M solution of a strong base
is closest to .

1. 2
2. 7
3. 9
4. 11
5. 13

4) Given that the H+ concentration of a solution
is 8.84*10^-5 M, calculate the pH of a solution.

1. 3.185
2. 3.487
3. 4.029
4. 4.041
5. 4.054
6. 4.066
7. 4.077
8. 4.405
9. 5.161
10. 9.541


5) What is [H3O+] when [OH-] = 3.3 * 10^-9?

3.3 * 10^-9 M
6.6 * 10^-5 M
1.0 * 10^-7 M
3.0 * 10^-6 M
3.3 * 10^-5 M

Hi jwheel!!


Question 1: Answer 2. 4.33 

If hydrosulfuric acid (H2S) is dissolved in water the following two
reactions occur:

H2S (aq) = H+ + HS-             K1, where K1 = 1.1*10^-7
HS- = H+ + S2-                  K2, where K2 = 1.0*10^-14


Consider the first reaction:
[H2S] = 0.02 - x (here we can aproximate this by disregarding the "-x"
which is very small in comparison because Ka1 = 1.1*10^-7)
So [H2S] = 0.02 
[H+] = [HS-] = x

Ka1 = [H+] * [HS-] / [H2S] = x^2 / 0.02 = 1.1*10^-7 
Then
x = sqrt(2.2*10^-9) = 4.69*10^-5

From the second reaction, we have
[HS-] = 4.69*10^-5 - y = 4.69*10^-5 (because Ka2 is a very small
number in comparison)
[H+] = 4.69*10^-5 + y = 4.69*10^-5
[S2-] = y.

Ka2 = [H+] * [S2-] / [HS-] = 4.69*10^-5 * y / 4.69*10^-5 = y

Then y = 1.0 * 10^-14

The total [H+] is x + y, then:
[H+] = 4.69*10^-5 + 1.0*10^-14 = 4.690000001 * 10^5
(as you can see here the second reaction will not affect the final pH
value, this is becaese Ka2 is very small in comparison with Ka1, you
can disregard the second reaction in diprotic acids where Ka1 >
1000*Ka2).

pH = -log (4.690000001 * 10^5) = 4.33

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

Question 2: Answer 6. 9.1 * 10^-6 M

When you put the salt in water you have the following dissociation:

NaH4CL + H2O --> NaH4+ + Cl- + H2O

Then the cations will dissociate:

NH4+ + H2O --> H3O+  + NH3  

Ka = 1*10^-14 / Kb = 1*10^-14 / 1.8*10^-5 = 5.56*10^-10


x  =  [H3O+]  =  [NH3]

[NH4+]  =  0.15 - x  = 0.15 (since Ka is small)


Ka = [H3O+] * [NH3] / [NH4+] = x^2 / 0.15 = 5.56*10^-10

x = sqrt(0.15 * 5.56*10^-10) = sqrt(8.33*10^-11) = 9.13*10^-6

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

Question 3: Answer 5. 13  

This is a strong base, so it will completely dissociate in water,
then:
[OH-] = 0.10 
pOH = -log(0.10) = 1
pH = 14 - pOH = 14 - 1 = 13

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

Question 4: Answer 5. 4.054 


"·Water is able to dissociate into hydronium (H3O+) and hydroxide ions
(OH-).
·The pH of a substance is a function of the proton concentration (or
H3O+ concentration) the substance. To determine the pH, the following
equation is used: pH = -log [H+] (or pH = -log [H3O+]).
·An acid is a proton (H+) donor, which increases the proton
concentration of a solution, therefore lowering the pH.
·A base is a proton acceptor, which brings down the proton
concentration of a solution, therefore bringing up the pH."
From "The Dissociation of Water and the pH scale" at Northland
Community & Technical College website:
http://www.northland.cc.mn.us/biology/Biology1111/Notes/dissociation.pdf


pH = -log [H+] = -log (8.84*10^-5) = 4.054

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

Question 5: Answer 3.0 * 10^-6 M


If [OH-] = 3.3 * 10^-9 M , then:

pOH = -log(3.3 * 10^-9) = 8.48

Since pH + pOH = 14 , then:

pH = 14 - pOH = 14 - 8.48 = 5.52

pH = -log [H3O+] ==> [H3O+] = 10^(-pH) = 10^(-5.52)
                            = 3.02*10^-6 

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

I hope this helps you, if you need a clarification, please post a
request for it, I will be glad to respond your requests.

Best regards.
livioflores-ga