A buffer is prepared by adding 355g NH3 MM1703 gmol to 7500

A buffer is prepared by adding 3.55g NH3 (MM=17.03 g/mol) to 750.0 mL of 0.175M of HCl. What is the pH of the buffer? (Kb = 1.8*10^-5) (NO VOLUME CHANGE)

Solution

we have:

Molarity of HCl = 0.175 M

Volume of HCl = 0.75 L

Molar mass of NH3 = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass of NH3 = 3.55 g

we have below equation to be used:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(3.55 g)/(17.034 g/mol)

= 0.2084 mol

mol of HCl = Molarity of HCl * Volume of HCl

mol of HCl = 0.175 M * 0.75 L = 0.1312 mol

We have:

mol of HCl = 0.1312 mol

mol of NH3 = 0.2084 mol

0.1312 mol of both will react

excess NH3 remaining = 0.0772 mol

Volume of Solution = 0.75 + 0 = 0.75 L

[NH3] = 0.0772 mol/0.75 L = 0.1029 M

[NH4+] = 0.1312 mol/0.75 L = 0.175 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

we have below equation to be used:

This is Henderson–Hasselbalch equation

pOH = pKb + log {[conjugate acid]/[base]}

= 4.745+ log {0.175/0.1029}

= 4.975

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.9755

= 9.02

Answer:  9.02