If pK, for CN at 25Â°C is 4.7, the pll of 0.5 M aqueousNaCN solution

1.	If pK, for CN at 25Â°C is 4.7, the pll of 0.5 M aqueousNaCN solution is:(a) 12c)11.5(b) 10(d) 11
Answer» NaCN ↔ Na++ CN- Here pKb = 4.7 ⇒ -logKb = 4.7 Hence Kb = 1.99 × 10-5 Now Kb= [Na][CN]/ [NaCN] As [NaCN] = 0.5 M [CN][Na] = 0.5 × 1.99 × 10-5 As [CN] = [Na] Therefore, [CN]2= 9.97 × 10-5 Hence the concentration of [CN-] = 3.15 × 10-3 Now pOH = -log{3.15 × 10-3} pOH = 2.5 Therefore pH = 14 - 2.5 = 11.49

If pK, for CN at 25Â°C is 4.7, the pll of 0.5 M aqueousNaCN solution is:(a) 12c)11.5(b) 10(d) 11

Answer»

NaCN ↔ Na++ CN-

Here pKb = 4.7

⇒ -logKb = 4.7

Hence Kb = 1.99 × 10-5

Now Kb= [Na][CN]/ [NaCN]

As [NaCN] = 0.5 M

[CN][Na] = 0.5 × 1.99 × 10-5

As [CN] = [Na]

Therefore, [CN]2= 9.97 × 10-5

Hence the concentration of [CN-] = 3.15 × 10-3

Now pOH = -log{3.15 × 10-3}

pOH = 2.5

Therefore pH = 14 - 2.5 = 11.49