Finding H3O+ (aq) ) from strong acid concentrations and OH- (aq) from strong base concentrations
2.
Finding H3O+
(aq)
and OH- (aq) using Kw.
Finding H3O+ (aq) from strong acid concentrations and OH- (aq) from strong base concentrations Step 1
Write the ionization equation (strong acid) or dissociation equation (strong base)
Step 2
Use the mole ratio to calculate the ion concentration in solution
Examples: 1. Find the H3O+
2.
in a 0.467 mol/L solution of hydrochloric acid.
Find the OH- (aq) in a 0.0356 mol/L magnesium hydroxide solution.
Exercises: 1. Find the H3O+
2.
(aq) )
(aq) )
in a 0.497 mol/L solution of nitric acid.
Find the OH- (aq) in a 0.0272 mol/L strontium hydroxide solution.
1
Chemistry
Name: ___________________________
Finding H3O+(aq) and OH-(aq) using Kw Background Information: Pure water has a very slight conductivity that can only be measured with very sensitive equipment, therefore water can form ions: H2O(l)
H+(aq)
+
OH-(aq)
Auto-ionization of water water molecules in a pure sample of water will collide with each other, some of which will react to produce ions: H2O(l) + H2O(l) H3O+ (aq) + OH-(aq)
reverse reaction is favored since the most prevalent species present is water reaction is at a maximum when there are no other chemical species present ie. in pure water
Kw, the Equilibrium Constant of Water
Figure: The autoionization of water
Also called ionization constant or ion-product constant
The expression for the equilibrium constant =
In pure water, the equilibrium concentration for both H3O+(aq) and OH-(aq) = 1.00 x 10-7 mol/L, at 250C.
Thus Kw = H3O+
(aq)
products reactants
OH- (aq) = 1.00 x 10-14 (mol/L)2
(Recall that water is not included in the expression since it is a liquid)
Addition of an acid or base will increase the H3O+ (aq) or OH- (aq) and shift the equilibrium in such a way as to reduce the concentration of the other ion.
Since Kw will not change with changes in concentration (unless there is a temperature change), the value of Kw can be used to find the unknown concentration of H3O+ (aq) or OH- (aq).
2
Chemistry
Name: ___________________________
Example: If 1.00 x 10-3 mol of NaOH(aq) is added to 1.00 L of water, at 250C, find the H3O+ Since H3O+
(aq)
(aq).
OH- (aq) = 1.00 x 10-14 (mol/L)2,
then H3O+(aq) = 1.00 x 10-14 (mol/L)2 =1.00 x 10-14 (mol/L)2 = 1.00 x 10-11 mol/L OH- (aq) 1.00 x 10-3 mol/L Thus addition of OH- to the water: - increased OH- (aq) from 1.00 x10-7 mol/L to 1.00 x 10-3 mol/L - decreased H3O+ (aq) from 1.00 x10-7 mol/L to 1.00 x 10-11mol/L Diagram to illustrate that the product of H3O+(aq) OH- (aq) is a constant H3O+
(aq)
OH- (aq)
= 10-1
10-4
10-7
10-10
10-13
= 10-13
10-10
10-7
10-4
10-1
1.
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2.
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Exercises:
Write the answers to the following questions in the blank spaces provided above.
1.
What is the product of H3O+
2.
Label each situation above as to whether the aqueous solution is acidic (A), basic (B) or neutral (N).
(aq)
OH- (aq) in each situation in the above figure?
3
Chemistry
Name: ___________________________
Conclusion: In any solution of an acid or a base both H3O+ (aq) and OH- (aq) are present but their concentrations will vary (and their product, at 25oC, will always be 1.00 x 10-14 mol/L) ACID: H3O+ In pure water,
(aq)
OH- (aq)
NEUTRAL: H3O+
BASE: (aq)
H3O+
(aq)
OH- (aq)
= OH- (aq)
Examples: 1.
Find the H3O+(aq) in a cleaning solution that is 0.00400 mol/L KOH.
2.
Find the OH- (aq) of a solution of apple juice in which H3O+(aq) = 6.0 x 10-3 mol/L.
3.
A recipe for lye soap calls for 3.40g of NaOH to be dissolved in water to produce a 15.0 mL NaOH solution. Find the OH- (aq) and H3O+(aq)
4
Chemistry
Name: ___________________________
Exercises: Assume standard conditions of 250C and 101 kPa for all questions below. 1.
Find the H3O+(aq) in milk of magnesia that has a OH- (aq) = 1.43 x 10-4mol/L.
2.
Find the OH- (aq) in lime juice where H3O+(aq) = 1.3 x 10-2 mol/L
3.
Find the OH- (aq) and H3O+(aq) in a solution of aluminum hydroxide prepared by dissolving 2.50g of the ionic compound in water to produce a 150.0 mL solution.
