Question

A buffer is created by combining 150.0 mL of 0.15M methylamine (CH3NH2) with 150.0 mL of 0.20M of its conjugate acid (CH,NH4+). The K, value for methylamine is 4.37 x 104. a) Calculate the pH of this buffer. b) How many grams of NaOH would need to be added to raise the pH of this buffer to 11.00?

Answer 1

Solution :

Part A) The pOH is calculated from Hendersen -Hasselbalch equation as,

pOH = pkb + log [Salt ] / [Base]

pkb is calculated as,

pkb = - log kb = - log 4.37 x 10^-4 = 4 - log 4.37

pkb = 4 - 0.64 = 3.36

Thus,

pOH = 3.36 + log (0.20 / 0.15)

pOH = 3.36 + 0.12 = 3.48

Hence,

pH = 14 - pOH = 14 - 3.48 = 10.52

Part B)

Addition of NaOH increases the concentration of base while decreases the concentration of salt.

Suppose concentration of NaOH = X mol/L

Thus,

For pH = 11, the pOH = 3

Therefore,

pOH = pkb + log (0.20 -X) / (0.15 +X)

3 = 3.48 + log (0.20 -X) / (0.15 + X)

log (0.20 -X) / (0.15+X) = - 0.48

log 0.20 - log X - log 0.15 - log X = - 0.48

log (0.20 / 0.15) - 2 log X = - 0.48

0.12 - 2 log X = - 0.48

2 log X = - 0.60

log X = - 0.30

X = antilog - 0.30 = 0.50 M

Hence, molarity of NaOH = 0.50 M

Since, total volume of buffer = 150 + 150 = 300 mL

Thus, using molarity equation.

Molarity = Number of moles / Volume in L

Number of moles = Molarity x Volume in L

= 0.50 M x 0.300 L = 0.15 mol

Hence,

Mass of NaOH = 0.15 mol x molar mass

= 0.15 M x 40 g mol-1 = 6.0 g of NaOH