Balance the ionic equation by half reaction method O3+ Br- -------> O2 + BrO-

Answer:-

Step -1  

       Find the oxidation number of Br in BrO-

       in combination state oxygen has oxidation number = - 2 

             Br + O = - 1 

               Br -2 = -1 

                   Br = +1  

Step -2 

        write oxidation half reaction 

               Br-  <-----------> BrO- 

               -1                       +1

       here oxidation number of Br is increasing by 2 

       so add 2 electron to balance it 

       we get 

              Br-  <-----------> BrO-   + 2e-             

Step -3 

add OH - ions to balance the charge reaction 

Br- +  2OH-   <----------->BrO-   + 2e

Step - 4 

add H2O to balance the hydrogen and oxygen 

Br- +  2OH-   <----------->BrO-   +H2O +  2e 

balanced oxidation half reaction

 Br- +  2OH-   <----------->BrO-   +H2O +  2e   ---------------(1)

similarly you can get balanced half reduction reaction 

Balanced reaction half reaction 

O3 + H2O + 2 e- <-----------> O2 + 2 OH-        ---------------(2)

add both equations we get 

Br- +  2OH-  + O3 + H2O + 2 e- <-----------> BrO-   + 2e + O2 + 2 OH-

simplify the equation we get

O3+  Br- <----------->O2 +  BrO-

Answer : -  Equation was already balanced 

Balance the ionic equation

how many grams are in 1.8 x 10^23 atoms of silver

number of moles of silver = number of atoms / Avogadro number 

                                       = 1.8x10^23 atom/ 6.022x10^23 per moles

                                       =0.2989 moles 

mass in grams of silver    = number of moles * molar mass of substance 

                                       = 0.2989 moles * 107.87 gram /moles 

                                       = 32.34 grams 

Answer = 32.34 grams 

how many grams are in 1.8 x 10^23 atoms of silver

What is the molarity of pure water at 20 C?density table of water

 molarity of pure water

   density of water is at 20 C = 0.9982071 gram /ml

let we take 1 liter water 

                                   1 liter = 1000 ml 

            mass of 1 liter water = volume *density  

                                             = 1000 ml * 0.9982071 gram /ml 

                                             =  998.2071 gram

  molar mass of water  H2O = 2* 1.008 + 15.999

                                             = 18.015 gram / moles

number of moles of water in one liter 

                                            = mass of substance / molar mass

                                            = 998.2071 gram/18.015 (gram / moles)

                                            =55.41 moles 

        molarity of water        = number of moles of water / volume of water 

                                           = 55.41 moles/ 1liter 

                                           = 55.41moles/liter

Answer:-  molarity of water at 20 C is 55.41M

What is the final pH after 1 drop (0.05 mL) of 6 M HCI is added to 1.0 L of freshly prepared pure water that was originally at a pH of 7.0. Is there a significant pH change?

What is the final pH after 1 drop (0.05 mL) of 6 M HCI is added to 1.0 L of freshly prepared pure water that was originally at a pH of 7.0. Is there a significant pH change?

Answer:- 

                added volume = 0.05 /1000  

                                        = 0.00005 liter

    number of add moles  = molarity * volume 

                                        = 6moles / liter  * 0.00005 liter 

                                        =0.0003 moles 

total volume of solution   = 1.0 liter  +  0.00005 liter 

                                         = 1.00005 liter 

molarity of new solution = number of moles / volume in liter 

                                        =0.0003/1.00005 

                                        = 2.99x10^-4

    pH is this new solution= - log [H+]

Reaction is HCl  ----------> H+   +   Cl -

                            so [H+] = 2.99x10^-4

                         hence PH =  - log 2.99x10^-4

                                          =  3.52 

Hence there are big change in pH value

Answer  :- yes there are significant change in pH value

what is the approximate expected pH of a 0.030 M HNO3 solution?

Answer:- 

 HNO3, HCl ,H2SO4 are the  strong acid .

The formula for strong acid is  

                                                 pH = - log [H+] 

    

Dissociation reaction of HNO3 is

                                          HNO3 ---> H+    + NO3-

             So concentration of [H+]  = 0.030M 

                                             pH = - log 0.030 

                                            pH = 1.52

Answer pH = 1.52

what is the approximate expected pH of a 0.030 M HNO3 solution?

 

A 0.2602 g sample of an unknown monoprotic acid requires 12.23 mL of 0.1298 M NaOH solution to reach the end point. what is the molecular mass of the acid?

Answer :-

number of moles of NaOH = molarity * volume in liter 

                                           =0.1298 moles / liter * (12.23/1000) liter  

divide by 1000 to convert ml to liter 

                                          =  0.001587454 moles

same of monoprotic acid so it will give one H + 

and base NaOH  gives only one OH - ions 

and net reaction will  H+   +    OH - ----> H2O 
 

 so moles of acid  = moles of base 

                             = 0.001587454 moles 

          molar mass = mass of substance / number of moles 

                             = 0.2602g/0.001587454 moles 

                             = 163.91 gram / moles 

 Answer :- molar mass of acid is 163.91 grams /mole

A 0.2602 g sample of an unknown monoprotic acid requires 12.23 mL of 0.1298 M NaOH solution to reach the end point. what is the molecular mass of the acid?

10.00 mL of vinegar (mass=10.05 g) requires 14.77 mL of 0.4926 M NaOH to reach the end point. Calculate the molarity and mass percent of the acetic acid in the vinegar.

 Solution:-

Calculation of molarity :-

                  at the end point base  M1*V1*N1 = M2*V2*N2

          molarity of vinegar acid (M1) = to find

                      volume of vinegar V1 = 10 ml 

       It gives one H + ion  so that N1= 1 

                   Molarity of NaOH M2 = 0.4926

                           volume of NaOH = 14.77 ml 

                 It gives one OH- so N2 = 1   

                                                                                    Now use the formula  

                                 M1*V1*N1 = M2*V2*N2 

                                                                                   plug the values in this formula we get 

                                    M1*10*1 = 0.4926*14.77*1 

                                         10 M1 = 7.275702

                                              M1 = 0.7276 M

                                                              Answer molarity of vinegar is 0.7276 M 

Calculation of mass % :-

number of moles of base  = Molarity * volume in liter  of base 

                                       =(0.4926 moles /liter)*(14.77ml /1000)    

                                                                                        divided by 1000 to convert volume ml to  liter

                                       = 0.007275702 moles

reaction between acid and base 

                         CH3COOH + NaOH ---> CH3COONa  + H2O 

                                                                  here one moles of acid is reacting with 1 moles of base 

so number of moles of acid = number of moles of base 

                                         =  0.007275702 moles 

 Molar mass of acetic acid  CH3COOH = 60.05 g / moles

Mass of vinegar is = molar mass * number of moles 

                            =60.05 g/moles  * 0.007275702 moles 

                           =  0.4369059051 gram 

grams % of vinegar =  mass of acetic acid * 100/ total mass vinegar 

                              = 0.4369059051*100/ 10.05 % 

                              = 4.35%

 Answer mass % of vinegar is 4.35%

10.00 mL of vinegar (mass=10.05 g) requires 14.77 mL of 0.4926 M NaOH to reach the end point. Calculate the molarity and mass percent of the acetic acid in the vinegar.

Q-The titration of 10.00 mL of a diprotic acid solution of unknown concentration requires 21.37 mL of a 0.1432 M NaOH solution. What is the concentration of the diprotic acid solution?

solution :-

The titration of 10.00 mL of a diprotic acid solution

we get volume of given diprotic acid (V1) = 10 ml 

diprotic acid always gives 2 H + ions 

                          so number of of H + (N1) = 2

                   molarity of diprotic acid (M1) = ?

21.37 mL of a 0.1432 M NaOH solution

NaOH gives only one OH ions 

                                           so value of N2 = 1 

                                                 Volume V2 = 21.37 ml

                                        and molarity M2 = 0.1432 M

Formula for the reaction   

                                               M1*V1* N1= M2*V2*N2

          plug all values we get 

                                        M1 * 10.00ml*2 = 0.1432 *21.37 ml * 1 

          solve this condition we get 

                                                             M1 = 0.1530 M 

Answer :-   molarity of diprotic acid is 0.1530 M 

Q-The titration of 10.00 mL of a diprotic acid solution of unknown concentration requires 21.37 mL of a 0.1432 M NaOH solution. What is the concentration of the diprotic acid solution?