CHAPTER 13.2-3 pH, pOH and the Ion Product of Water
1. b able to both accpet and donate a proton.
2. a H+ and b OH-
3. d 1.0x10e-14 = [H+] x [OH-]
4. a the dissociation of water into its ions
5. neutral
acidic
basic
basic
6. a increases
7. c 2e-7 M
8. b decreases
9. b basic
10. a acidic
11. b lower, acidic
12. basic
acidic
neutral
13. c 11.0
14. d the conjugate acid of water in an acid is ionization
15.d HNO3(aq) + H2O --> [H3O+](aq) + [NO3-](aq)
16. b pH=1
17. a [Na+]=0.50 M, [OH-]=0.50 M, and [H+]=2.0e-14 M
18. d Use a stoichiometric calculation
*Mr. Henderson pointed out that we should write down the equations/formulas on the right side of the page on a notecard*
equations/formulas:
- [H3O+] x [OH-] = 1.0e-14
- [H+] x [OH-] = 1.0e-14
- pH= -log[H3O+]
- pH= -log[H+]
- [H3O+]=[H+]=10^(-pH)
- pOH= -log[OH-]
- [OH-]= 10^(-pOH)
- pH + pOH = 14 (at 25 degrees C)
We then turned to page #4 where we worked on #9 part a.
a. the first step is to write the equation
H2O + HCN <---> [H3O+] + [CN-]
I 2.0 0 0
C -x +x +x
E 2-x x x
Ka expression= ([CN-] x [H3O+])/[HCN]
Ka=6.2e-10=(x^2)/(2-x)=(x^2)/(2)-->(x^2)=1.24e-9
x=3.5e-5
This is when the 5% approximation comes into play. This says that since the value of x is so small that it can just be ignored in the 2-x part of the equation.
3.5e-5 M=[CN-]=[H3O+]
We then turned to page #6 to finish the table in probblem 12 and work on question #13
12. [H3O+] [OH-] pH pOH acidic or basic?
1.0e-2 1.0e-13 2.00 12.00 acidic
1.0e-11 1.0e-3 11.00 3.00 basic
1.0e-5 1.0e-9 5.00 9.00 acidic
1.0e-3 1.0e-11 3.00 11.00 acidic
1.0e-7 1.0e-7 7.00 7.00 neutral
2.0e-14 5.0e-2 13.70 0.30 basic
6.0 1.67e-15 -0.78 14.78 acidic
2.5e-15 4.0 14.60 -0.60 basic
13. a. [H+]= 2.0 M
pH= -log(2.0)=-0.30
b. [OH-]= 2.0 M
pOH= -log(2.0)= -0.30
pH= 14.30
c. [OH-]= 0.100 M
pOH= -log(0.100)=1.00-->pH=13.00
Review of the six strong basses:Li, Na, K, Ca, Sr, Ba
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