The solubilituy of `PbCI_(2)` in water is `0.01M 25^(@)C`. Its maximum concentration in `0.1M NaCI` will be:

The solubility of PbCl_(2) in water is 0.01 M 25^(@) C. Its maximum concentration in 0.1 M NaCl will be :

The solubility of PbCl_2 in water is 0.01 M at 25^@C it maximum concentration in 0.1M NaCl will be :

If the maximum concentration of PbCl_2 in water is 0.01 M at 25^@C , its maximum concentration in 0.1 M NaCl will be

The solubility product of AgCI in water is 1.5xx10^(-10) . Calculate its solubility in 0.01M NaCI .

Calculate the change in pH when a 0.1 M solution of CH_3COOH in water at 25^@C is diluted to a final concentration of 0.01 M . [K_a=1.85xx10^(-5)]

If S_(0), S_(1), S_(2) , and S_(3) are the solubility of AgCI in water, 0.01M CaCI_(2), 0.01M NaCI 1, and 0.5M AgNO_(3) solutions, respectively, then which of the following is true ?

The degree of ionisation of an acid HA is 0.00001 at 0.1 M concentration. Its dissociation constant will be

The degree of ionisation of an acid HA is 0.00001 at 0.1 M concentration. Its dissociation constant will be

A 0.01 M solution of glucose in water freezes at - 0.0186^(@)C . A 0.01 M solution of NaCl in water will freeze at