## 6th Edition, 12.23

Acidity $K_{a}$
Basicity $K_{b}$
The Conjugate Seesaw $K_{a}\times K_{b}=K_{w}$

Douglas Nguyen 2J
Posts: 71
Joined: Fri Sep 28, 2018 12:15 am

### 6th Edition, 12.23

12.23: The value of Kw for water at body temperature (37 C) is 2.1*10^(-14) mol*L-1.
(a) What is the molarity of H3O+ ions and the pH of neutral water at 37 C?

I calculated the pH, and I got 6.838890353; when rounding, I should get 6.83, but the textbook says the answer is 6.80.

Can someone please explain the reasoning behind 6.80 being the pH?

KatelinTanjuaquio 1L
Posts: 69
Joined: Fri Sep 28, 2018 12:16 am
Been upvoted: 1 time

### Re: 6th Edition, 12.23

I got the same answer as you, 6.83 and I am unsure why the book says 6.80.I think it is a calculation error.

In the solutions manual, it states that square root of 2.1x10^-14 is 1.4x10^-7, when I calculated 1.45x10^7. Just to test their answer, I plugged in their value, 1.4x10^-7, to the pH formula. -log[H3O+]=-log(1.4x10^-7) and that came out to be 6.85, which is very different from their given value of pH, 6.80.

Because the values are inconsistent, I think it is a calculation error.

Lynsea_Southwick_2K
Posts: 55
Joined: Fri Sep 28, 2018 12:25 am

### Re: 6th Edition, 12.23

Maybe because of sig figs its 6.8? It is weird that the book states 6.80

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