Question

1) You have an aqueous solution where [OH-] = 1 x 10-4 mol/L.

Answer 1

1. A. 1 x$10^{-10}$x  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) $K_{w}$= [$H_{3}$O+] [OH]-     $K_{w}$= 1×$10^{-14}$

  [$H_{3}$O+]= 1×$10^{-14}$ ÷ 1 x 10-4

             = 1 x$10^{-10}$x  is the hydrogen ion concentration.

     

B) pH =-log [$H_{3}$O+]

pH = -log[-1x $10^{-10}$]

   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 = $\frac{number of moles}{volume}$  

Number of moles can be calculated as:

n = $\frac{mass}{atomic mass of one mole of substance}$

n = $\frac{195}{58.44}$    ( 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 = $\frac{number of moles}{volume}$

M   =   $\frac{3.33}{2.8}$

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×  $\frac{152}{1000}$  =  M2 × $\frac{750}{1000}$

M2 = $\frac{0.75}{0.456}$

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  $\frac{20}{1000}$ = Macid x $\frac{31}{1000}$

0.0048 = Macid x 0.031

Macid= $\frac{0.031}{0.048}$

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