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We have, the concentration of  and the volume of the solution containing sulphur ion = 10 mL. Volume of metal salts solution added = 5mL Before mixing, M  After mixing, Volume = 15 mL So, the concentration of                                                       concentration of                                             So, the ionic product =                                         ...
We have, The solubility product of calcium sulphate is . given mass of calcium sulphate = 1g Ionization of calcium sulphate; Therefore,  Let the solubility of calcium sulphate be . Then,        mol/L Thus, mass/ (mol. wt) volume  Molarity mass =  So, that to dissolve 1 g of calcium sulphate we need =  of water.
We have, The solubility product of the  Equals number of moles of ferrous sulphate and sodium sulphide are mixed in an equal volume. Let  be the concentration of ferrous sulphate and sodium sulphide. On mixing the equimolar solution, the volume of the concentration becomes half. So,  The ionisation of ferrous sulphide; Therefore, for no precipitation, ionic product = solubility product By...
It is given that,   of buffer solution is 3.19. So, the concentration of  ion can be calculated as;               = antilog (-3.19)              =  Ionization of benzoic acid; It is given that  Therefore, Let the   
We have, solubility product () of cupric iodate = When equal volumes of sodium iodate and cupric chlorate are mixed together the molar concentration of both the solution becomes half (= 0.001) Ionization of cupric iodate is;  0.001 M                                   0.001 M So,  can be calculated as;             Sinc eionic product is less than the  so no precipitation occurs.
silver chromate () Ionization of silver chromate Let "" be the solubility of      of  =        Ionization of Silver bromide ()   of  =          Now, the ratio of solubilities              
Solubility product is the product of ionic concentrations in a saturated solution. (i) silver chromate () Ionization of silver chromate Let "" be the solubility of    According to the table  of  =        (ii) Barium chromate () Ionization of silver chromate Let "" be the solubility of    According to the table  of  =        (iii) Ferric hydroxide () Ionization of Ferric...
Given that, Volume of 0.1 M  = 10 mL, and  Volume of 0.1 M  = 10 mL So, by using the formula of, By putting the values we get, Hence, 
In this case, both the solutions have the same number of moles of and , therefore they both can get completely neutralised. Hence the   = 7.0
Given that, Vol. of 0.2 M  = 10 mL Vol. of 0.1 M HCl = 25 mL  therefore, by using the formula, By substituting the value in these equations, we get; Now,                        since                                = 14-1.221                    =   12.78                    
We have the ionic product of water at 310 K is  It is known that, ionic product  SInce , therefore   at 310 K is   here we can calculate the value of  concentration. Thus,                                         Hence the  of neutral water is 6.78
We have, Ionisation constant of chloroacetic acid() is  The concentration of acid = 0.1 M Ionisation if acid, =  We know that, ....................(i) As it completely ionised Putting the values in eq (i) Therefore,  of the solution =                                                    =                                                    =  Now,   0.1 M  (sod. chloroacetate) is basic...
Salts of strong acid and strong base are neutral in nature for example-  Salts of a strong base and weak acid are basic in nature for example- Salts of strong acid and a weak base are acidic in nature for example-
Given,     =  3.44 We know that By taking antilog on both sides we get,  = antilog (- 3.44) pyridinium hydrochloride completely ionised. Then  =  (conc. of products)/ (conc, of reactants)                  =        (? Concentration is 0.02M) Now,    (approx)
We have, Ionization constant of nitrous acid =  Concentration of sodium nitrite () = 0.04 M Degree of hydrolysis can be calculated as; Sodium nitrite is a salt of sodium hydroxide (strong base) and the weak acid () Suppose  moles of salt undergoes hydrolysis, then the concentration of- , and  Therefore  from here we can calculate the value of  ;                                    ...
We have, Concentration of cyanic acid = 0.1 M   Therefore, the concentration of  = antilog (-2.34)                                                             =  It is known that,    =    Then Ionization constant ()  =                                                                                                           
Let the degree of ionization of propanoic acid be . Then Let suppose we can write propanoic acid to be HA,   It is known that, We have  ionization constant of propanoic acid ()=  and the concentration is 0.005 M     By putting the values in above formula we get, [Acid] =  = C. =  Therefore,                                                      If we add 0.01M hydrochloric acid then,  ...
By given abova data, we know the solubility of  at 298 K = 19.23 g/L So, concentration of      =  (Molecular weight of  = 121.63 u) = 0.1581 M and the concentration of  Now It is known that,              =  Therefore 
We have 0.562 g of potassium hydroxide (). On dissolving in water gives 200 mL of solution. Therefore, concentration of =                                                                      = 2.805 g/L                                                                                                                                             It is a strong base. So, that it goes complete...
We already know that  can be calculated as-   to calculate the concentration of  = antilog (-) Thus, the hydrogen ion concentration of followings  values are- (i)  of milk = 6.8 Since,  6.8 =   = -6.8  = anitlog(-6.8)         =   (ii) of black coffee = 5.0 Since,  5.0 = = -5.0 = anitlog(-5.0) =   (iii)  of tomato juice = 4.2 Since, 4.2 = = -4.2 = anitlog(-4.2) =   (iv)  of lemon juice =...
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