**Sample Problem 1 **: What are the concentrations of hydronium and hydroxide ions in a beverage whose pH =3.05 ?

Solution:

**Step1** : To convert from pH to ion concentrations,first apply equation 17-1 to calculate [H3O+]. Then make use of water equilibrium to calculate [OH-]

**Step 2 **: We must rearrange equation pH = -log [H+] , in order to solve for concentration

**Step 3** : log [H+] = -pH and [H+] = 10 ^{-pH }

**Step 4** ; Now substitute and evaluate:

[H+] = 10^{-3.05 }= 8.9 x 10^{-4} M

**Step 5**: To determine the hydroxide ion concentration,we need to rearrange the water equilibrium expression:

Kw = 1.0 x 10 ^{-14 }= [H3O+]_{eq} [OH-] and [OH^{-}]_{eq} = 1.0 x 10^{- 14} / [H3O^{+}]

now susbtitute and evaluate:

= 1.0 x 10^{- 14 }/ 8.9 x 10^{-4} = 1.1 x 10^{-11} M

**Sample protblem 2** : What is the OH- ion concetraion of a solution of NaOH with pH 13.40 ?

Solution:

**Step 1** : First find pOH from the given pH:

pH + pOH = 14

pOH = 14 - 13.40 = 0.60

**Step 2** : As you have now the value of pOH you can find the OH- ion concentration using formula

pOH = -log [OH-]

[OH^{- }] = 10 ^{- pOH }= 10 ^{- 0.60 }= 0.25 M