`50.0` mL of `0.10` M ammonia solution is treated with `25.0` mL of `0.10M HCI`. If `K_(b)(NH_(3))=1.77xx10^(-5)`, the pH of the resulting solution will be

50 mL of 0.2M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_(b) of ammonia solution is 4.75 the pH of the mixture will be :

50 mL of 0.2 M ammonia solution is treated with 25mL of 0.2 M HCl . If pK_(b) of ammonia solution is 4.75 , the pH of the mixture will be

pH of when 50mL of 0.10 M ammonia solution is treated with 50 mL of 0.05 M HCI solution :- (pK_(b) "of ammonia"=4.74)

pH of when 50mL of 0.10 M ammonia solution is treated with 50 mL of 0.05 M HCI solution :- (pK_(b) "of ammonia"=4.74)

Calculate the pH after 50.0mL of this solution is treated with 25.0mL of 0.1M HCI K_(b) for NH_(3) = 1.77 xx 10^(-5) (pK_(b) ~~ 4.76) .

Calculate the pH after 50.0mL of 0.1M NH_(3) this solution is treated with 25.0mL of 0.1M HCI K_(b) for NH_(3) = 1.77 xx 10^(-5) (pK_(b) ~~ 4.76) .

50 mL of 0.1 M HCl and 50 mL of 2.0 M NaOH are mixed. The pH of the resulting solution is

Calculate the pH after 50.0 mL of 0.1 M ammonia solution is treated with 25.0 mL of 0.10 M HCl. The dissociation constant of ammonia , K_b=1.77xx10^(-5)

At 25^(@) C, K_b for CH_3COO^(0-) ion in water is 5.55xx10^(-10) If 100mL of 0.025(M) HCl solution is added to 50mL of 0.1(M) solution of CH_3COONa , pH of the resulting solution will be-

10mL of 0.1 M HCl solution reacts completely with 10 mL of 0.1 M NaOH solution. What will be the pH of the resulting solution?