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7.55     Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

(d)     Human saliva, 6.4.

we have  6.4 It is known that  Therefore, By taking antilog on both sides we get,

7.55     Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

(c)       Human blood, 7.38

we have  7.38 It is known that  Therefore, By taking antilog on both sides we get,

7.55     Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

(b)        Human stomach fluid, 1.2

We have  1.2 It is known that  Therefore, By taking antilog on both sides we get,

7.55     Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

(a)     Human muscle-fluid, 6.83

We have 6.83 It is known that  Therefore, By taking antilog on both sides we get,

7.54     The ionization constant of dimethylamine is 5.4 × 10-4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?

We have, (Degree of ionization)  Concentration of dimethylamine = 0.02 M       If we add 0.1 M of sodium hydroxide. It is a strong base so, it goes complete ionization                                                                       (0.1 M)      (0.1 M) and also,     0.02-                                                                        (since the dissociation is very...

7.53     Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains

(b)     0.1M in HCl ?

Let the  amount of acetic acid is dissociated in this case                                   Initial conc.                  0.05                     0                              0 after dissociation        0.05 -                0.1+                  As the dissociation is very small. So we can write 0.1+   0.1 and  0.05 -    0.05                      So, the value of  =  Now the degree of...

7.53     Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains

(a)     0.01M

We have, C = 0.05 M By taking antilog on both sides we get, from here we get the value of                                                         After adding hydrochloric acid, the concentration of  ions increases and due to that the equilibrium shifts towards the backward direction. It means dissociation will decrease. (i) when 0.01 HCl is taken                              Initial...

7.52     What is the pH of 0.001M aniline solution?  Calculate the degree of ionization of aniline in the solution. Also, calculate the ionization constant of the conjugate acid of aniline.(K= $4.27\times 10^{-10}$)

We have, C = 0.001 M  =  Degree of inozation of aniline (a) = ? Ionization constant of the conjugate acid () = ? We know that     =         =  (0.001) Thus                  =  Then [Base] = C.a = ()()                                =  Now,               = 14 - 6.187            = 7.813 It is known that, So,                    This is ionization constant.

7.51     The pH of 0.005M codeine $\left ( C_{18}H_{21}NO_{3} \right )$ solution is 9.95. Calculate its ionization constant and pKb.

We have, C = 0.005 M  = 9.95 and  = 14 - 9.95 = 4.05  we know that                          By taking antilog on both sides we get. concentration of  =  C.a =  So, a =  We know that,          Thus

7.50     The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.

We have, Degree of ionization(a) = 0.132  Concentration of bromoacetic acid (C) =  0.1 M Thus the concentration of                                                               = 0.0132  Therefore                                      = 1.879  Now, we know that,  So,                    (approx)

7.49     Calculate the pH of the following solutions:

(d)      1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

We know that, M1V1(before dilution) = M2V2(after dilution) initially V1 = 1mL and M1 = 13.6 M and V2 = 1L and M2 = ? By putting all these values we get, Thus                                                 = 1.86 (approx)

7.49     Calculate the pH of the following solutions:

(c)     0.3 g of NaOH dissolved in water to give 200 mL of solution.

dissociates into  So, the concentration of   =    We know that , Now,  =

7.49     Calculate the pH of the following solutions:

(b)      0.3 g of $Ca(OH)_2$ dissolved in water to give 500 mL of solution.

The calcium hydroxide ion dissociates into- Molecular weight of  = 74  the concentration of  =    We know that,                                   =  Thus                                                 = 14 - 1.79  = 12.21

7.49     Calculate the pH of the following solutions:

(a)     2 g of TlOH dissolved in water to give 2 litre of solution.

Here, 2 g of  dissolves in water to give 2 litres of solution So, the concentration of    =       (the molar mass of   is 221)  can be dissociated as  Therefore,             (since Kw = ) So, the concentration of  =  Thus                                                 = 11.65(approx)

7.48     Assuming complete dissociation, calculate the pH of the following solutions:

(a)      0.003 M HCl

(b)      0.005 M NaOH

(c)      0.002 M HBr

(d)       0.002 M KOH

Assuming the complete dissociation. So,  (a) The ionisation of hydrochloric acid is  Since it is fully ionised then  Therefore,   of the solution =                                  =                                  = 2.52  (b) The ionisation of     Therefore,  of the solution =                                       of the solution is equal to (14 - 2.301 =11.70)  (c)  The ionisation...

7.47     It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.

We have, pH of organic acid is 4.15 and its concentration is 0.01M Suppose the organic acid be HA. The dissociation of organic acid can be written as; (By taking antilog of -4.15) Now, [HA]  = 0.01 Then, Thus

7.46     The ionization constant of acetic acid is 1.74 × 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

It is given, The ionisation constant of acetic acid is  and concentration is 0.05 M The ionisation of acetic acid is; Therefore,      So, the  of the solution =                                              =                                             = We know that,

7.45   The first ionization constant of H2S is 9.1 × 10-8. Calculate the concentration of HS- ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H2S is 1.2 × 10-13, calculate the concentration of S2- under both conditions.

We have, 1st ionisation constant of hydrogen sulphide is  and the 2nd dissociation constant is  Case 1st-(absence of hydrochloric acid) To calculate the concentration of  Let  be the concentration of and the ionisation of hydrogen sulphide is;  0.1 M At equilibrium, the concentration of various species are, Since the dissociation constant is is very small. So,  can be neglected. the...

7.44     The ionization constant of phenol is 1.0 × 10-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?

We have,  The ionization constant of phenol is , and the concentration of phenol is 0.05 M degree of ionisation = ? Ionization of phenol; At equilibrium,  the concentration of various species are- As we see, the value of ionisation is very less. Also  will be very small. Thus we can ignore . Hence the concentration of phenolate ion is  Let  be the degree of dissociation of phenol in...

7.43     The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10-4, 1.8 × 10-4 and 4.8 × 10-9 respectively. Calculate the ionization constants of the corresponding conjugate base.

We have, IOnization constant of hydrogen fluoride, methanoic acid and hydrogen cyanide are  and  respectively. It is known that, ...........................(i)  of the conjugate base     Similarly,  By using the equation (i)  of the conjugate base   Again, with the help of eq (i)  of the conjugate base
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