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a buffer is created by combining 150.0 mL of 0.15M methylamine (CH3NH3) with 150.0 mL of...

a buffer is created by combining 150.0 mL of 0.15M methylamine (CH3NH3) with 150.0 mL of 0.20 M of its conjugate base (CH3NH4+). The Kb value for methylamine is 4.37 x 10^-4.
A) how many grams of NaOH would need to be added to raise the pH of this buffer to 11.00?
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Answer #1

A)

moles of CH3NH2 = 0.150 x 0.15 = 0.0225

moles of CH3NH3+ = 0.150 x 0.20 = 0.030

Kb = 4.37 x 10^-4

pKb = 3.36

pKa = 10.64

pH = pKa + log [base + C / acid - C]

11.00 = 10.64 + log [0.0225 + C / 0.030 - C]

[0.0225 + C / 0.030 - C] = 2.29

0.0225 + C = 0.0687 - 2.29 C

C = 0.0136

moles of NaOH = 0.0136

mass of NaOH added = 0.545 g

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