Please read the paragraph and answer the questions asked
In chemistry, pH is a numeric scale used to specify the acidity or alkalinity of an aqueous solution. It is the negative of the logarithm to base 10 of the activity of the hydrogen ion. Solutions with a pH less than 7 are acidic and solutions with a pH greater than 7 are alkaline or basic.
(i)calculate the pH of the following solution 100 ml of 0.1 M acetic acid is mixed up with 10 ml of 0.1 M hcl (ka for acetic acid=1.8*10^-5)
(ii)calculate pH of 10^-8 M of NaOH.
(iii)what will be the pH of the following mixture .1000 ml of solution which is 0.1 M wrt HCL and 0.05M wrt Naoh
(iv)calculate ph of 0.1 m HCL and 40 ml 0.2 (M) h2 so4

1
answers for 1 st part is 2.04
2nd part is 7.02
3rd part is 1.3

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2015-10-19T05:39:34+05:30

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1)
seems that the 10 ml of HCl is not negligible as compared to 100 ml of Acetic acid C2 H3OOH.

100 ml of Acetic acid = 0.1 M  ie., has 0.01 moles.
So 110 ml of solutions has 0.01 moles.
so it's concentration is now  0.01 *1000/110  M = 0.0909 M

concentration of [H+] from HCl : 
   10 ml of 0.1 M HCl contains :  0.001 moles
   concentration of [H+] in 110 ml of solution : (HCl dissociates completely)
            0.001 * 1000/110 =  0.00909

Acetic acid will dissociate into ions very little..  because Ka is very small.  So we assume that [H+] in the solution is nearly:  0.00909 M
               pH = - log 0.00909 = 2.0414

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2)    NaOH solution in water..  As the concentration of the base is very small of the order (in fact less than) that of water, we have to consider that also.  Let us consider 1 litre of solution.

 The  10^-8 M of Na OH contain 10^-8 moles of  [OH-]  ions.  From water suppose x moles of  [OH-] and x  moles of [H+] are dissociated.
 
   So total concentration of  [OH-] ions =  (x + 10^-8) M
 
   for water:    [ H3O+] [ OH-] = 1 * 10^-14
                    x (x + 10^-8) = 10^-14
                   x² + 10^-8 x  - 10^-14 = 0
                   let x = a * 10^-7

                     a² + 0.1 a  - 1 = 0
                       a = { -0.1 + - √[0.1²+4] } / 2
                       a   ≈  0.95
             so  [H+]  0.95 * 10^-7
                   pH =  7.022

================
3)  
  1 liter:  0.1 M HCl  and    0.05 M of Na OH
  so there are 0.1 Moles of [H+] and  0.05 Moles of [OH-] in the solution.
  Then 0.05 Moles combine to form water.
   so 0.05 moles of [H+] remain in the solution.
   pH =  - log 0.05 = 1.301
===========================
4)
     0.1 M  HCl  and  40 ml of  0.2 M  H2 SO4.
   I assume that 1 liter of HCl is taken.   If we ignore change in the volume due to H2SO4 addition, then it is simpler.  If we do not do that, then we have to recalculate the concentrations.

   0.1 M  HCl has now concentration = 0.1 M * 1000/1040 = 0.0961 M
   0.2 M H2SO4 has now concentration = 0.2 M * 40/1040  =0.00769
     H2SO4 gives two H+ ions.  so the final concentration of [H+] will be:
                    0.0961 + 2 * 0.00769 = 0.1115 M
   pH = - log [H+] = 0.9525

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