Ionic EquilibriumHard
Question
To prepare a buffer of pH 8.26 amount of (NH4)2 SO4 to be added to 500 mL of 0.01 M NH4OH solution [pKa (NH4+) = 9.26] is
Options
A.0.05 mole
B.0.025 mole
C.0.10 mole
D.0.005 mole
Solution
For the buffer solution of NH3 & NH4+
pH = pKa + log
⇒ 8.26 = 9.26 + log 
⇒ m. moles of NH4+ = 50 ∴ moles of (NH4)2 SO4 required = 0.025.
pH = pKa + log
⇒ 8.26 = 9.26 + log ⇒ m. moles of NH4+ = 50 ∴ moles of (NH4)2 SO4 required = 0.025.
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
Solution of aniline hydrochloride is X due to hydrolysis of Y. X and Y are :...The conjugate base of H2PO4- is...What is the pH of 10−7 M-HCl solution at 25o C?...100 ml of pH = 6 (acidic) is diluted to 1000 ml by H2O. pH will increase by :-...The solubility of AgCl in water is 0.001435 g per litre at 15o C. The solubility product of AgCl is(Ag = 108, Cl = 35.3)...