Practice set test , grade yourself – complete all problems
1-1. 10-10 joules = 0.1 nJ or 100pJ
(a) 4.317 28 X 10 -8 coulombs = 43.1728 nC
(b) 2.997 9 X 1014 hertz = 299.19 THz
(c) 10-10 meter 100 pm or 0.1 nm
(d) 2.1 X 10 13 watts 21 TW
(e) 48.3 X 10-20 moles = 483 zmol
1-2.
(a) ppm
 1g  1  1L
 1mL  6
 10  0.0282 ppm
(10.7 g ) 6 


3
 10 g  0.345L  110 mL  1.10 g 
 
(b) ppb = 1000*ppm, so it is 28.2 ppb
1-3.
(a) mass HClO4 = (1000.0 g soln)(0.705 g solute/1000 g soln) = 705 g
(b) mass water = 1000.0 g soln – 705 g solute = 295 g
(c) mole HClO4 = (mass HClO4)/(FM=100.4329 g/mole)=7.01 mole
1-4.

 100 g so ln 
NaOH 
40.00 g / mole
  32 g solution
1L

 12.5 g NaOH 
1.00L 0.10mol

1-5
 110 6 g solute  1.08 g  103 mL 

  1.08  10 3 g / L
Conc  

 g solution  mL  L 
 1106 g 
  1.08  103 g / L
Conc  1.08 10 3 g / L 
1
g


 1L

Conc  1.08 103 g / L 
  1.08g / mL
3
 110 mL 




 1103 mg 
  1.08mg / L
Conc  1.08 10 3 g / L 
g




1-6.
concentration of C20H42 is:
 0.35 109 g solute 


1

 1103 g solution / L 
  1.2 109 M
g solution
 282.55 g / mole 




1-7.
(a) Conc of CsCl:
 40.0 g CsCl  143g solution  1mole CsCl 


  3.397 M or 3.40M
Conc  
 100 g solution  0.1000 L  168.36 g 
(b) Volume of the concentrated solution needed:
Vol  0.100M 0.800 L  / 3.397 M   0.0236 L or 23.6mL(also could be 24mL due to 800mL having 1 sig fig )
1-8.
TD glassware is calibrated to deliver a known volume of liquid at a given temperature that is printed on
the glassware. However, TC glassware is calibrated to contain a known volume at a given temperature that is
printed on the glassware. The typical temperature for calibration is 20 °C.
1-9.
The volume is just the mass multiplied by the volume of water/g at the specified temperature, 26 ºC (see
Table 2-7:
Vol  5.000 g 1.0043mL / g   5.0215 or 5.022mL (4 sig figs)
1-10. Clearly, the TD glassware needs to be calibrated, which is what you will do in Chem 2002 laboratory.
The true volume is:
Vol= m/d =m (1/d)
Vol = (15.569g ) * (1.0040 mL/g see Table 2-7 ) = 15.631 mL (5 SF)
1-11.
You are to make a 1.500 M solution of Cu(NO3)2 in water. Assuming you will make 0.60000 L of this solution,
how would you prepare it? And I mean exactly how would you prepare it?
The mass of Cu(NO3)2 needs to be determined first:
Mole =M * V , mole= Mass/MM,
Mass = M* V * MM
mass  0.60000L1.500M 187.5558 g / mole  168.80 or 168.8g
Due to the fact that the amount of salt is so large and that its dissolution in water with cause the temperature of
the solution to change, the following procedure is followed (this is the route one would usually follow for any
solution preparation):
a. weigh out 168.8 g of the copper salt;
b. dissolve the salt in a clean beaker containing roughly 300 mL of pure water by stirring with a glass stirring
rod;
c. let the temperature of the solution equilibrate to RoomTemp (do not remove the stirring rod);
d. quantitatively transfer the contents of the beaker to a 600.00 mL volumetric flask (to do so, you would rinse
the beaker with a squirt bottle of pure water and then pour the rinses into the volumetric flask using the stirring
rod as a liquid guide; the beaker should be rinsed roughly 4 times with ~5 mL of water and the rinses
transferred to the VF using the stirring rod; finally, the stirring rod should be rinsed into the VF with ~5 mL of
pure water.);
e. QS to the mark with pure water (this means that you add a quantity sufficient of water to bring the volume of
the water in the VF to the scribed mark; the bottom of the meniscus should be at the line);
f. mix by inversion (hold your finger on the stopper!) 10-25 times to ensure proper mixing and to avoid errors in
sampling of the top layer of the VF contents.
1-12. First of all, this is called a serial dilution. In this case, we have made two 1:50 dilutions, which is a total
1:2500 dilution. Thus, the concentration of the copper nitrate solution is (1.500 M)/(2.5x103) = 6.000 x
10-4 M. You may also do the calculations in steps, which would lead to 0.03000 M after the first
dilution and 0.0006000 M after the second dilution. There should be 4 sig figs in the final answer.
Now, the concentration of nitrate is twice that of the Cu(II) or 1.200 x 10 -3 M for nitrate and 6.000 x 10-4 M for
Cu(II). This is due to the electrostatics of the situations (no excess charge can be floating around in solution).
1-13.
This is easy. Just take the base-10 log of the hydrogen ion concentration for this strong acid. Thus, the pH is 4
(you only have 1 sig fig here).
1-14.
Okay, this is not as easy as the previous problem. The HOAC is a weak acid, so it does not completely ionize
upon being placed in water. So, the HOAc is in equilibrium with its conjugate base, acetate (OAc-) as follows:
(a) HOAc <----> OAc- + H+; Ka = 1.75 x 10-5
Now, the formal concentration means that all species containing acetate (HOAc and OAC-) should sum up to
1.0000 moles/L.
(b) For every mole of HOAc that does ionize, an equivalent amount of the conjugate base and hydronium ion
(or solvated proton) is formed. That is to say that the equilibrium concentration of the acetic acid is less than
the initial 1.0000 M by an amount that is equal to the concentration of proton produced. Thus:
2
H  OAc 
H H
Ka 

 1.75 x 105 or H   H  K a  K a FHoac  0

HOAc  FHOAc  H
 
   
 
   
We needed to obtain an equation that has only one unknown so that we could solve it. PLEASE, DO NOT USE
X for your variables, as later on we will learn that this can be confusing and lead to errors.
Now, we can use the quadratic formula to find the hydrogen ion concentration:
H 

 K a  K a  4(1)(  K a FHAOc )  1.75 x 10 5 


2(1)(1)
2
1.75 x 10 
 4(1)(1.75 x 10 5 )1.0000 
2(1)(1)
5 2
We of course ignore the negative root, as that would give us a negative proton concentration. Thus, we find that
the hydrogen ion concentration is 4.175 x 10-3 M. When we take the base-10 log of this, we obtain (with 3 sig
figs for the mantissa) that the pH is 2.379 for this weak acid solution. Typically, we will only be able to
measure the pH to the hundreths place with most pH meters, so if you put down 2.38 for the pH, that is fine.
Now, you would say to me, why did you use the QF to find the hydrogen ion concentration? Well, you could
use the approximation that the formal concentration of the HOAc does not change significantly upon its
ionization so that the equilibrium concentration is effectively the formal concentration. That is true in this case,
but if you made that approximation, you MUST show that such an approximation is appropriate. Now, how do
you make sure that you can do so? Well, if the ratio of the resulting hydrogen ion concentration to the formal
concentration of the species of interest (HOAc) is less than 1%, then you can use the approximation. So, if one
uses the approximation, the hydrogen ion concentration is the square root of the product of the formal HOAc
conce and the Ka value, which is 4.18 x 10-3 M. When one determines the %ratio, it is 0.4%. Thus, the
approximation is valid. BUT, later on we will use formal concentrations of weak acids/bases that are such that
we cannot make this approximation.