1) You have an aqueous solution where [OH-] = 1 x 10-4 mol/L.
A. Calculate the hydrogen ion concentration [H+]. Show all work.
B. Calculate the pH. Show all work.
C. Is this solution an acid, base, or neutral? You must explain how you know.
2) Calculate the molarity of a solution prepared by dissolving 195 g of sodium chloride (NaCl) in enough water to make 2.8 liters of solution.
3) A student takes 152 mL of a 3.0M HCl solution and places it into a beaker. The student then adds enough water to the beaker to make 750 mL. Calculate the molarity of this new solution. Show all work.
4) A student started a titration by placing 20.0 mL of 0.24 M NaOH into a flask with 3 drops of phenolphthalein. The student then titrated to the endpoint using 31.0 mL of HCl. Calculate the concentration (molarity) of the HCl that the student used. Show all work for credit.

1. A. 1 xx  is the hydrogen ion concentration.

 B. pH of the solution is 10

  C. The solution is basic.

2. The molarity of NaCl in 2.8 litres of water is 1.18 M

3. 1.64 M is the molarity of new solution.

4. 0.64 M is the molarity of the acid  or HCl used.

Explanation:

1. Data given [OH-] =1 x 10-4   mol/L.

A) = [O+] [OH]-     = 1×

  [O+]= 1× ÷ 1 x 10-4

             = 1 xx  is the hydrogen ion concentration.

B) pH =-log [O+]

pH = -log[-1x ]

   pH  = 10

c) pH value of 10 indicates that it is a basic solution because it is greater than 7 and on pH scale more than 7 value indicates basicity.

2. Molarity is calculated by the formula

Molarity =  

Number of moles can be calculated as:

n =

n =    ( atomic weight of NaCl is 58.44 gm/mole)

n= 3.33 moles

Now the molarity of NaCl in 2.8 litres of water is calculated as:

Molarity =

M   =  

M = 1.18

the molarity of NaCl in 2.8 litres of water is 1.18 M

3. Data given:

Initial volume of the HCl solution V1 = 152 ml, initial molarity M1 = 3

final volume of the diluted HCl V2= 750 ml, final molarity M2= Unknown

The initial and final volumes are converted into litres in calculation.

So, M1V1 = M2V2

   3×    =  M2 ×

M2 =

M2 = 1.64 M is the molarity of new solution.

4. In titration the formula used is: M acid x V acid = M base x V base

Volume of base V1= 20 ml Molarity of base M1 = 0.24 M

volume of the acid V2 = 31 ml Molarity of the acid = unknown

the volume will be converted to litres.

So applying the formula,

M acid x V acid = M base x V base

0.24 x   = Macid x

0.0048 = Macid x 0.031

Macid=

M base= 0.64 M is the molarity of the acid  or HCl used.

the answer is true. hope it .

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.