## Value of Kw

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

hojae_lee_1C
Posts: 51
Joined: Fri Jun 23, 2017 11:39 am

### Value of Kw

The value of Kw for water at body temperature (37°C) is 2.1 X 10^-14. (a)What is the molarity of H3O+ ions and the pH of neutral water at 37°C? (b) What is the molarity of OH- in neutral water at 37"C?

How would I approach part (a) and (b)?

Sarah_Wilen
Posts: 62
Joined: Fri Jun 23, 2017 11:39 am
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### Re: Value of Kw

The value of Kw for water at body temperature (37°C) is 2.1 X 10^-14. (a)What is the molarity of H3O+ ions and the pH of neutral water at 37°C? (b) What is the molarity of OH- in neutral water at 37"C?

How would I approach part (a) and (b)?

Notice that the value of Kw shifted from 1.0 x 10^-14 at 25°C to its new Kw at 37"C: 2.1 X 10^-14. Now that we've noticed it, we can continue.

1) I start by writing out the equation of water's dissociation in water.
$H_{2}O_{(l)}+H_{2}O_{(l)}\rightleftharpoons H_{3}O^{+}_{(aq)}+OH^{-}_{(aq)}$

2) I write out my Kw equation and solve for the concentrations of hydronium and hydroxide by squaring both sides.
$K_{w}=[H_{3}O^{+}][OH^{-}]$
$2.1\times 10^{-14}=x^{2}$
$\sqrt{2.1\times 10^{-14}}=\sqrt{x^{2}}$
$1.4\times 10^{-7}M=[H_{3}O^{+}]=[OH^{-}]$

3) Now, we solve for the pH just how we always would. (negative log)
$pH=-log(1.4\times 10^{-7})=6.8$

yay