A chemistry graduate student is given 250. mL of a 1.00 M trimethylamine CCH, N solution. Trimethylamine is a weak base with Kb=7.4x 10 . what mass of (CH3)3NHBr should the student dissolve in the (CH3)3N solution to turn it into a buffer with pH-11.467 You may assume that the volume of the solution doesn't change when the (CH3) NHBr is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.

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Mergus Mergus · Answer 1 · 2019-11-11T08:16:29+01:00

Answer:

8.8 g

Explanation:

Given:-

[Trimethylamine] = 1.00 M

Volume = 250 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 0.25 L

Given that:

[tex]K_{b}=7.4\times 10^{-4}[/tex]

[tex]pK_{b}=-log\ K_{b}=-log\ (7.4\times 10^{-4})=3.13[/tex]

pH = 11.467

Also, pH + pOH = 14

So, pOH = 14 - 11.467 = 2.533

Considering the Henderson- Hasselbalch equation for the calculation of the pOH of the basic buffer solution as:

[tex] pOH=pK_b+log\frac{[salt]}{[base]} [/tex]

So,

[tex] 2.533=3.13+log\frac{[salt]}{1.00} [/tex]

[Salt] = 0.2526 M

Considering:

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

So,,

[tex]Moles =Molarity \times {Volume\ of\ the\ solution}[/tex]

So, Moles of [tex](CH_3)_3NHBr[/tex] = 0.2526*0.25 moles = 0.06315 moles

Molar mass of [tex](CH_3)_3NHBr[/tex] = 140.02 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]0.06315\ mole= \frac{Mass}{140.02\ g/mol}[/tex]

[tex]Mass_{(CH_3)_3NHBr}= 8.8\ g[/tex]