Ionic solid `Na^(+)A^(-)` crystallise in rock salt type structure.2.6gm of ionic solid salt NaA dissolved in water to make 2 litre solution.The pH of this solution is 8. If distance between cation and anion is 300pm calculate density of ionic solid (in gm/`cm^(3)`). (Given: `pK_(w)=13`
`pK_(a)``(HA)=5` `N_(A)=6.022times10^(23)`

Ionic solid Na^(+) A^(-) crystallise in rock salt type structure . 2.592 gm of ionic solid salt NaA dissolved in water to make 2 litre solution . The pH of this solution is 8. If distance between cation and anion is 300 pm. Calculate density of ionic solid (P low =13,P low (HA)=13)

Ionic solid B^+A^(-) crystallizes in rock salt type of structure.1.296 gm ionic solid salt B^+A^(-) is dissolved in water to make one litre solution.The pH of the solution is measured to be 6.0.if the value of face diagonal in the unit cell of B^+A^(-) be 600sqrt2 pm.Calculate the density of ionic solid in gm/cc. (T=298 K, K_b "for" BOH "is" 10^(-6) ,(Avogadro Number = 6.0xx10^(23) ))

The crystal AB (rock salt structure) has molecular weight 6.023 y u, where y is an arbitrary number in u. If the minimum distance between cation and anion is y^(1//2) nm and the observed density is 20kg/ m^(3) , find the (i) density in kg/ m^(3) and (ii) type of defect.

How many gm of solid KOH must be added to 100 mL of a buffer solution to make the pH of solution 6.0, if it is 0.1 M each w.r.t. acid HA and salt K A. [Given : pK_(a)(HA)=5]

A solid cube of edge length =25.32 mm of an ionic compound which has NaCl type lattice is added to 1kg of water.The boiling point of this solution is found to be 100.52^@C (assume 100% ionisation of ionic compound).If radius of anion of ionic solid is 200 pm then calculate radius of cation of solid in pm (picometer). ( K_b of water =0.52 K kg "mole"^(-1) ,Avogadro's number, N_A=6xx10^(23),(root3(75))=4.22 )

5.35 g of a salt Acl (of weak base AOH) is dissolved in 250 ml of solution. The pH of the resultant solution was found to be 4.827. Find the ionic radius of A^(+) and Cl^(-) if ACl forms CsCl type crystal having 2.2 g/cm3. Given K_(b(AOH))=1.8xx10^(-5), (r^(+))/(r^(-))=0.732 for this cell unit.

Mentallic Gold crystallise in fcc lattice and the length of cubic unit cell is 407 pm. (Given : Atomic mass of Gold = 197 ,N_(A)=6xx10^(23) ) The density if it have 0.2% Schottky defect is (" in gm"//cm^(3))

A sample of hydrazine sulphate (N_(2)H_(6)SO_(4)) was dissolved in 100 mL water. 10 mL of this solution was reacted with excess of FeCl_(3) solution and warmed to complete the reaction. Ferrous ions formed were estimated and it required 20 mL of M//50 KMnO_(4) solutions. Estimate the amount of hudrazine sulphate in one litre of solution. Given 4Fe^(3+)+N_(2)H_(4)rarrN_(2)+4Fe^(2+)+4H^(+) MnO_(4)^(-)+5Fe^(2+)+8H^(+)rarrMn^(2+)+5Fe^(3+)+4H_(2)O

A solution which remains in equilibrium with undissolved solute , in contact , is said to be saturated . The concentration of a saturated solution at a given temperature is a called solubility . The product of concentration of ions in a saturated solution of an electrolyte at a given temperature is called solubility product (K_(sp)) . For the electrolyte A_(x),B_(y) with solubility S. The solubility product (K_(sp)) is given as K_(sp) = x^(x) xx y^(y) xx S^(x-y) . While calculating the solubility of a sparingly . soluable salt in the presence of some strong electrolyte containing a common ion , the common ion concentration is practically equal to that of strong electrolyte containing a common ion . the common ion soncentration is practically equal to that of strong electrolyte . If in a solution , the ionic product of an electroylte exceeds its K_(sp) value at a particular temperature , then precipitation occurs . If two or more electrolyte are presentt in the solution , then by the addition of some suitable reagent , precipitation generally occurs in increasing order of their k_(sp) values . Solubility of some sparingly soluable salts , is sometimes enhanced through complexation . While we are calculating the solubility of some sparingly or pH of an electrolyte , the nature of cation of anion should be checked carefully whether there ion (s) are capable of undergoing hydrolysis or not . If either or both of the ions are capable of undergoing hydrolysis , it should be taken into account in calculating the solubility . While calculating the pH of an amphiprotic species , it should be checked whether or not cation can undergo hydrolysis . Total a_(H^(-)) = sqrt(K_(a_(1)xxK_(a_(2)))) (if cation do not undergo hydrolysis ) a_(H^(+)) = sqrt(K_(a_(1))((K_(w))/(K_(b)) - K_(a_(2)))) (if cation also undergoes hydrolysis ) where symbols have usual meaning . Solubility of solids into liquids is a function of temperature alone but solubility of gases into liquids is a function of temperature as well as pressure . The effect of pressure on solubility of gases into liquids is governed by Henry's law . The solubility of BaSO_(4) in 0.1 M BaCl_(2) solution is (K_(sp) " of " BaSO_(4) = 1.5 xx 10^(-9))

Calculate the equilibrium constants for the reactions with water of H_(2)PO_(4)^(Theta), HPO_(4)^(2-) , and PO_(4)^(3-) as ase. Comparing the relative values of two equilibrium constants of H_(2)PO_(4)^(Theta) with water, deduce whether solutions of this ion in water are acidic or base, Deduce whether solutions of HPO_(4)^(2-) are acidic or bases. Given K_(1),K_(2) ,and K_(3) for H_(3)PO_(4) are 7.1 xx 10^(-3), 6.3 xx 10^(-8) , and 4.5 xx 10^(-13) respectively.