Question

You are asked to prepare 1.1L of a HCN/NaCN buffer that has a pH of 9.77 and an osmotic pressure of 1.97atm at 298 K.

What masses of HCN and NaCN should you use to prepare the buffer? (Assume complete dissociation of NaCN.)

......... I put in ( 1.09 ) and ( 1.09 ) but it says its incorrect,,,,, Please Help!!

Answer 1

we know that

for a buffer

pH = pKa + log [conjugate base / acid ]

so

pH = pKa + log [NaCN / HCN]

pKa for HCN is 9.21

so

9.77 = 9.21 + log [NaCN / HCN]

[NaCN / HCN] = 3.63078

[NaCN] = 3.63078 [HCN]

given

osmotic pressure = 1.97

we know that

osmotic pressure = i x c x R x T

NaCN --> Na+ + CN-

so

i = 2

now

osmotic pressure = i x c x R x T

1.97 = 2 x C x 0.0821 x 298

C= 0.04 M

so

the conc of the soltuion is 0.04 M

so

total molarity = 0.04 M

[HCN] + [NaCN] = 0.04

but

[NaCN] = 3.63078 [HCN]

so

[HCN] + 3.63078 [HCN] = 0.04

[HCN] = 8.694 x 10-3

so

[NaCN] = 3.63078 x 8.694 x 10-3

[NaCN] = 0.031566 M

given 1.1 L

we know that

moles = molarity x volume

so

moles of HCN = 8.694 x 10-3 x 1.1 = 9.56 x 10-3

also

mass = moles x molar mass

so

mass of HCN = 9.56 x 10-3 x 27

mass of HCN = 0.258

so

0.258 g of HCN is required

similarly

moles of NaCN = 0.031566 x 1.1 = 0.0347226

mass of NaCN = 0.0347226 x 49

mass of NaCN = 1.70 g

so

0.258 grams of HCN and 1.70 g of NaCN is required