Question

Question 16 A chemistry graduate student is given of a methylamine solution. Methylamine is a weak base with . What mass of should the student dissolve in the solution to turn it into a buffer with pH ? You may assume that the volume of the solution doesn't change when the is dissolved in it. Be sure your answer has a unit symbol, and round it to significant digits.

Answer 1

Answer:

90 g CH₃NH₃Cl

Explanation:

It appears your question lacks the values required to solve this problem. However, an online search tells me these are the values. Be aware if your values are different your answer will also be different, but the methodology remains the same.

" A chemistry graduate student is given 450 mL of a 1.70 M methylamine solution. Methylamine is a weak base with Kb=4.4x10⁻⁴. What mass of CH₃NH₃Cl should the student dissolve in the methylamine solution to turn it into a buffer with pH = 10.40 ? You may assume that the volume of the solution doesn't change when the is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits. "

We'll use the Henderson-Hasselbach equation:

• pH = pKa + log$\frac{[Methylamine]}{[Salt]}$

So first we use Kb to calculate Ka and then pKa:

• Ka = Kw/Kb ⇒ Ka = 1x10⁻¹⁴/4.4x10⁻⁴ = 2.27x10⁻¹¹

• pKa = -log(Ka) = 10.64

Now we can calculate the concentration of the salt, CH₃NH₃Cl:

• pH = pKa + log$\frac{[Methylamine]}{[Salt]}$

• 10.40 = 10.64 + log$\frac{1.70}{[Salt]}$

• -0.24 = log$\frac{1.70}{[Salt]}$

• $10^{-0.24}$=$\frac{1.70}{[Salt]}$

• [Salt] = 2.95 M

Now we use the final volume to calculate the moles of CH₃NH₃Cl and finally its mass, using its molecular weight:

• 450 mL ⇒ 450/1000 = 0.450 L

• 2.95 M * 0.450 L = 1.3275 mol CH₃NH₃Cl

• 1.3275 mol CH₃NH₃Cl * 67.45 g/mol = 89.54 g CH₃NH₃Cl

Which when rounded to 2 significant digits becomes 90 g CH₃NH₃Cl