HW Problem 6D. 17

Moderators: Chem_Mod, Chem_Admin

Gisela F Ramirez 2H
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

HW Problem 6D. 17

Postby Gisela F Ramirez 2H » Wed Jan 23, 2019 11:05 pm

Calculate the pH of:
a.) 0.63 M NaCH3CO2
b.) 0.65 M KCN

How exactly do I go about working out this problem without a K value? I have been looking at the 6D.1 table and nothing seems to match so I am not sure how to solve it

Anusha 1H
Posts: 65
Joined: Fri Sep 28, 2018 12:15 am

Re: HW Problem 6D. 17

Postby Anusha 1H » Wed Jan 23, 2019 11:16 pm

For this question, we actually have to refer to table 6C.1 which the acidity constants.
In the case of a), you know that you have to work with CH3CO2-, so we can use the Ka value of acetic acid (CH3COOH) to find the Kb value of its conjugate base.
Since Ka for acetic acid is 1.8 x 10^-5, the Kb value should be 5.62 x 10^-10. With this, you just set it up like the other problems we've been working with by using an ice table.

For b), its pretty much the same thing but use the Ka value hydrocyanic acid (HCN) to find the Kb value of its conjugate base.

Tuong-Minh Tran 1C
Posts: 30
Joined: Fri Sep 28, 2018 12:19 am

Re: HW Problem 6D. 17

Postby Tuong-Minh Tran 1C » Wed Jan 23, 2019 11:24 pm

Since the problem does not give you a K value, you have to derive it yourself. While Table 6D.1 does not have what you need, Table 6C.1 (page 461) does. To clarify:

a) NaCH3CO2 is basically acetic acid with a sodium ion attached. Therefore, you refer to the Ka value of 1.8*10^-5 (pKa=4.75).

b) Similarly, think of KCN as CN- with K+ attached. The table gives the Ka value of the conjugate acid HCN, which is 4.9*10^-10 (pKa=9.31).

With these values, you should be able to find K.


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 0 guests