Question

Determine the mass of NH₄Cl and the volume of 0.250 M NaOH solution required to make 0.15 L of a buffer solution at pH=9.20 and at a total concentration of 0.42 M. pK_a(NH₄⁺)=9.25.

1) Mass of NH₄Cl:

2) Volume of NaOH solution:

Answer 1

Mass of NH4Cl = 3.3699g

Volume of NaOH solution = 118.8ml

Explanation

Henderson - Hasselbalch equation is

pH = pKa + log([NH3/[NH4+])

9.20 = 9.25 + log([NH3/[NH4+])

log([NH3]/[NH4+]) = -0.05

[NH3]/[NH4+] = 1*10^-0.05=0.891

[NH3] = 0.891[NH4+]

Total concentration of buffer = 0.42M

[ NH3 ] + [ NH4+ ] = 0.42M

0.891[NH4+] + [NH4+] = 0.42M

1.891[NH4+] = 0.42M

[NH4+] = 0.222M

[ NH3 ] = 0.420M - 0.222M = 0.198M

So, Concentraion requiremets are

[ NH3]= 0.198M

[ NH4+ ] =0.222M

Now , let us see how to achive these concentration for 0.15L

NH4+ react with NaoH to give NH3

NH4+ + OH^- -------> NH3 + H2O

this reaction is 1:1 molar reaction,so

If 0.42 mol of NH4Cl is allowed to react with 0.198mol of NaOH,

0.198 mole of NH3 will form, 0.222mole of NH4+ will remain

the above calculation is for 1000ml

for 150ml

mole of NH4+ =( 150/1000)*0.42mol= 0.063mol

mole of NaOH = (150/1000)*0.198 = 0.0297mol

Molar mass of NH4+ = 53.49g/mol

mass of NH4Cl required = 0.063mol*53.49g/mol= 3.3699g

molarity of NaOH solution = 0.250M

Volume of NaOH Solution required =( 1000ml/250mol)*0.0297mol = 118.8ml