acid base chemistry

Question

What quantity in moles of CH₃NH₃Cl need to be added to 200.0 mL of a 0.500 M solution of CH₃NH₂ (Kb for CH₃NH₂ is 4.4 × 10⁻⁴) to make a buffer with a pH of 11.45?

Answer 1

What quantity in moles of CH₃NH₃Cl need to be added to 200.0 mL of a 0.500 M solution of CH₃NH₂ (Kb for CH₃NH₂ is 4.4 × 10⁻⁴) to make a buffer with a pH of 11.45?

You can use another form of the Henderson Hasselbalch equation (for basic buffers).

pOH = pKb + log [salt] / [base]

pOH = 14 - pH = 14 - 11.45 = 2.55

pKb = -log Kb = 3.36

[salt] = [CH₃NH₃Cl] = ?

[base] = 0.500 M

Solving for [salt], we have ...

2.55 = 3.36 + log [salt] / [base]

log [salt] / [base] = -0.81

[salt] / [base] = 0.155

[salt] / 0.500 = 0.155

[salt] = 0.0774 M

Moles of salt (CH₃NH₃Cl) needed = 0.0774 mol / L x 0.200 L = 0.0155 moles

Answer 2

given the pH and the Kb we can use the henderson hasselbach equation to find the ratio of the moles of acid to moles of base

hederson hasselbach equation: pH=pKa + log [base]/[acid]

pH= 11.45

in order to find the pKa we need to convert from Kb to Ka and then from Ka to pKa

Ka*Kb=1.0E-14 Ka= 1E-14/4.4E-4 Ka=2.27 E-11 pKa=-logKa = 10.644

plugging back into the equation we get: 11.45=10.644 + log (base/acid)

base/acid=6.4.054

we know moles of base= (0.5M)(0.2L)=0.1

0.1 moles base/X moles acid= 6.397

moles acid=0.0156 moles