Textbook 6b1

Moderators: Chem_Mod, Chem_Admin

Ethan Goode 2H
Posts: 122
Joined: Wed Sep 30, 2020 9:51 pm

Textbook 6b1

Postby Ethan Goode 2H » Sat Dec 12, 2020 3:28 pm

6B.1 The molar concentration of HCl in hydrochloric acid is reduced to 12% of its initial value by dilution. What is the difference in the pH values of the two solutions?
How do you do this problem? I am confused where to start.

Stuti Pradhan 2J
Posts: 153
Joined: Wed Sep 30, 2020 9:32 pm
Been upvoted: 5 times

Re: Textbook 6b1

Postby Stuti Pradhan 2J » Sat Dec 12, 2020 3:51 pm

The easiest way to do this problem is to simply choose a molar concentration of HCl and find the pH. Then, take 12% of that molar concentration and find the pH value. Subtract the values to find the difference in pH.

For example, I used a 0.5 M solution of HCl, and then a 0.06 M solution (which is 12% of 0.5M), found the two pHs and then found the difference between them.

Hope this helps!

Catie Donohue 2K
Posts: 91
Joined: Wed Sep 30, 2020 9:55 pm

Re: Textbook 6b1

Postby Catie Donohue 2K » Sat Dec 12, 2020 3:55 pm

I couldn't find the solution for this in the textbook - would the correct answer be ~0.921?

Sofia Lombardo 2C
Posts: 85
Joined: Wed Sep 30, 2020 9:31 pm

Re: Textbook 6b1

Postby Sofia Lombardo 2C » Sat Dec 12, 2020 4:47 pm

The way I approached this problem was to use . So 0.12 [HCl] - [HCl] and then I took the -log of this value and divided by the initial [HCL] (because it is property of log functions) and then the [HCl] would cancel and I was left with just -log(0.12).

Eric Cruz 2G
Posts: 82
Joined: Wed Sep 30, 2020 9:45 pm
Been upvoted: 1 time

Re: Textbook 6b1

Postby Eric Cruz 2G » Sat Dec 12, 2020 5:33 pm

For this problem I used 1M concentration as the original. Therefore, if you put it in the -log(H) equation, you get 0. 12% of 1M is 0.12M, so I put 0.12 in the -log(H) equation as well and got me 0.92 . I took the difference of the two which is merely just 0.92

Return to “Calculating pH or pOH for Strong & Weak Acids & Bases”

Who is online

Users browsing this forum: No registered users and 1 guest