Ionisation of water, pH and buffers

Question 1

keq is what and what is its formula

Answer 1

Equilibrium constant and Keq = (products)/(reactants)

For H2O Keq = (H+)(OH-) / (H2O)

Question 2

What is Kw

Answer 2

the ionic product.

For H2O kw = (H+)(OH-) = 10^-14

Question 3

Calculate Keq for H2O given that the concentration of water is 70

Answer 3

Keq = (H+)(OH-) / (H2O)
kw = (H+)(OH-) = 10^-14
Keq= 10^-14 / 70                         (10^-7) = H+   (10^-7) = OH-

10^-14 / 70 = 1.4 x 10^16
Keq = 1.4 x 10^16

Question 4

What is Ka and what is its formula

Answer 4

Ka = (H+) (Conjugate base) / (Reactant)

So the formula for ka of ascetic acid (CH3COOH) would be

Ka = (H+) (CH3COO - ) / (CH3COOH)

Question 5

Using Ka to find the acid dissociation

Answer 5

Ka of ascetic acid = 1.76 x 10^-5
To find the dissociation of 20mM = 2x10^-2 M of CH3COOH we must find H+ concentration
CH3COOH = H+ + CH3COO
1.76 x 10^-5 - x (x) (x)
so
1.76 x 10^-5 x (2x10^-2) = (x) (x)
(x) (x) = 3.5 x 10^-7
x = SQRT( 3.5 x 10^-7)
x = 5.9 x 10^-4
So out of 20mM only 0.6 mM of acid dissociated

Question 6

Using Ka to find the Ph

Answer 6

pH = -log (H+)
H+ = 5.9 x 10^-4
-log (H+) = 3.2 which makes sense cuz its a ascetic acid is a weak acid
Ph or Ka can be found using either the given Ka or pH and concentration of acid or base

Question 7

What is a buffer

Answer 7

A substance that can mitigate ph change of an added acid or base

Question 8

Henderson Hasselbach equation and what is it used for?

Answer 8

Ph = pKa + Log(acid/conj base)
used to calculate the ph of a solution

If the acid and its conj base are in equilibrium (same concentration -log(1) = 0 which means that at equal concentrations the ph and Pka are the same