Ionic EquilibriumHard
Question
Ka for HCN is 5 × 10-10 at 25oC. For maintaining a constant pH of 9, the volume of 5 M KCN solution required to be added to 10 ml of 2M HCN solution is (log 2 = 0.3)
Options
A.4 ml
B.8 ml
C.2 ml
D.10 ml
Solution
Ka = 5 × 10-10 pKa = 10 log 5 = 9.3
pH = pKb + log
9 = 9.3 + log
⇒ - 0.3 = log 
0.3 = log
⇒ 
= 2 ⇒ Vml = 2 ml
pH = pKb + log
9 = 9.3 + log
⇒ - 0.3 = log 0.3 = log
⇒ = 2 ⇒ Vml = 2 mlCreate a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
Some chemists at ISRO wished to prepare a saturated solution of a silver compound and they wanted it to have the highest...A weak acid HA after treatment with 12 mL of 0.1 M strong base BOH has a pH of 5. At the end point the volume of same ba...Which one of the following is NOT a buffer solution?...The hydronium ion concentration in an aqueous solution of H2SO4 is 2.0 × 10−4 M at 25o C. The hydroxide ion concentratio...The $pK_a$ of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized...