Question

A chemistry graduate student is given of a dimethylamine solution. Dimethylamine 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

Due to incomplete data in the question, the exact mass calculation cannot be performed. However, the method involves using the Henderson-Hasselbalch equation and the molarity of the dimethylamine solution to find the mass of the conjugate acid needed to create a buffer with a specific pH.

Step-by-step explanation:

The question seems to be missing some numerical data, such as the pKa value of dimethylamine and the desired pH for the buffer solution. However, assuming we had this information, typically, you would use the Henderson-Hasselbalch equation, which is used for buffer calculations:

PH = pKa + log10([A-]/[HA])
Here, [A-] represents the concentration of the conjugate base of dimethylamine (CH3NH2) and [HA] represents the concentration of the weak base dimethylamine itself. The student would need to calculate the amount of the conjugate acid (often in the form of a chloride salt like CH3NH3Cl) to add to their dimethylamine solution to achieve the desired pH.

First, rearrange the equation to solve for [A-]/[HA]. Then, using the molarity of the dimethylamine solution, calculate the mass of the conjugate acid to add. Remember to take the molar mass of the conjugate acid into account.

Answer 2

This is an incomplete question, here is a complete question.

A chemistry graduate student is given 500 mL of a 0.20 M pyridine C₅H₅N solution. Pyridine is a weak base with Kb = 1.7 × 10⁻⁹. What mass of C₅H₅NHCl should the student dissolve in the C₅H₅N solution to turn it into a buffer with pH = 4.76 ?

You may assume that the volume of the solution doesn't change when the C₅H₅NHCl is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.

Answer : The mass of C₅H₅NHCl is, 34 grams.

Explanation :

First we have to calculate the value of
`pK_b`.

The expression used for the calculation of
`pK_b` is,

`pK_b=-\log (K_b)`

Now put the value of
`K_b` in this expression, we get:

`pK_b=-\log (1.7* 10^(-9))`

`pK_b=8.77`

Now we have to calculate the value of
`pK_a`

`pK_a+pK_b=pK_w\\\\pK_a+8.77=14\\\\pK_a=5.23`

Now we have to calculate the concentration of C₅H₅NHCl.

Using Henderson Hesselbach equation :

`pH=pK_a+\log ([Salt])/([Acid])`

`pH=pK_a+\log ([C_5H_5N])/([C_5H_5NHCl])`

Now put all the given values in this expression, we get:

`4.76=5.23+\log ((0.20)/([C_5H_5NHCl]))`

`[C_5H_5NHCl]=0.59M`

Now we have to calculate the mass of C₅H₅NHCl.

`\text{Concentration of }C_5H_5NHCl=\frac{\text{Mass of }C_5H_5NHCl* 1000}{\text{Molar mass of }C_5H_5NHCl* \text{Volume of solution (in mL)}}`

Now put all the given values in this formula, we get:

`0.59=\frac{\text{Molar mass of }C_5H_5NHCl* 1000}{115.5g/mole* 500mL}`

`\text{Molar mass of }C_5H_5NHCl=34.0725g\approx 34g`

Thus, the mass of C₅H₅NHCl is, 34 grams.