500 মিলিমিটার দ্রবণে 4.9 গ্রাম সালফিউরিক অ্যাসিড থাকে। সমাধানের pH গণনা করুন। ...

Calculate the amount of (NH_(4))_(2)SO_(4) in grams which must be added to 500ml of 0.2MNH_(3) to give a solution of pH=9.3 .Given pK_(b) for NH_(3)=4.7 .

Calculate the weight of (NH_(4))_(2)SO_(4) which must be added to 500mL of 0.2M NH_(3) to yield a solution of pH = 9.35. K_(a) for NH_(3) = 1.78 xx 10^(-5) .

What will be the amount of (NH_(4))_(2)SO_(4) (in g) which must be added to 500 mL of 0.2 M NH_(4)OH to yield a solution of pH 9.35? ["Given," pK_(a)"of "NH_(4)^(+)=9.26,pK_(b)NH_(4)OH=14-pK_(a)(NH_(4)^(+))]

Calculate the amount of (NH_(4))_(2)SO_(4) in grams which must be added to 500 ml of 0.2 M NH_(3) to yield a solution of pH=9 , K_(b) for NH_(3)=2xx10^(-5)

Calculate the pH of a solution which results from the mixing of 50.0 ml of 0.3 M HCl with 50.0 ml of 0.4 M NH_(3). [K_b (NH_(3)0=1.8 xx 10^(-5)]