30.4k views
0 votes
Calculate the pH of a 0.324 M (CH3)2NH₂Br solution. The Kb of (CH3)₂NH is 5.9 x 10-4.

Enter your answer with three decimal places.

User Akhlesh
by
5.9k points

1 Answer

3 votes

Answer:

Hope this helps ;) don't forget to rate this answer !

Step-by-step explanation:

To calculate the pH of a solution, we need to know the concentration of the hydronium ions, [H+]. The concentration of hydronium ions is related to the concentration of the conjugate base of a weak acid through the equilibrium expression:

[H+][A-]/[HA] = Kb

In this case, the weak acid is (CH3)2NH2 and its conjugate base is (CH3)2NHB. The concentration of the conjugate base, [A-], is equal to the concentration of the salt (CH3)2NHBr, which is 0.324 M. The Kb of (CH3)2NH is 5.9 x 10-4.

Substituting these values into the equilibrium expression, we get:

[H+][(CH3)2NHB]/[(CH3)2NH2] = 5.9 x 10-4

We can rearrange this equation to solve for [H+]:

[H+] = Kb * [HA]/[A-]

= 5.9 x 10-4 * [HA]/0.324

We don't know the concentration of the weak acid, [HA], so we can't solve for [H+] directly. However, we can use the fact that the sum of the concentrations of the weak acid and its conjugate base must be equal to the concentration of the salt, 0.324 M.

[HA] + [A-] = 0.324

Substituting the concentration of the conjugate base, [A-], into this equation, we get:

[HA] + 0.324 = 0.324

[HA] = 0

Since [HA] is equal to zero, the concentration of the weak acid is also zero. Substituting this value into the equation for [H+], we get:

[H+] = 5.

User HenryAdamsJr
by
6.5k points