# Note to Parents

```Acid and Base Problems
1.
0.1 mol of HCl is added to 1 L of water.
What is [H+] and [OH-]?
H2O  H+ + OHHCl  H+ + Cl[HCl](dissolved) = [H+] = 0.1mol/L = 1 x 10-1 M
[H+]= 1 x 10-1 M
Kw = [OH-][H+] = 1.0 x 10-14
[OH-] = 1.0 x 10-14/[H+]
= 1.0 x 10-14/[1 x 10-1]
= 1.0 x 10-13 M
So, [H+] = 10-1 M, [OH-] = 10-13 M
2.
0.04 g of NaOH(s) is added to 200 mL of water. What is [OH-]?
NaOH  Na+ + OHChange grams to moles first: 0.04 x (22.98977 + 15.9994 + 1.0079) = 0.04/39.99707 = 1.0 x
10-3 mol
Change mL to L: 200/1000 = 0.2 L
[NaOH](dissolved) = [OH-] = 1.0 x 10-3 mol/0.2 L = 0.005 M = 5.0 x 10-3 M
3.
A beaker contains 100 mL of water. We add 112 mg of KOH(s). What is the concentration of H+
ions?
KOH  K+ + OHChange milligrams to grams: 112/1000 = 0.112 g
Change grams to moles: 0.112/(39.0983 + 15.9994 + 1.0079) = 0.112/56.1056 = 0.001996236
mol = 1.996236 x 10-3 mol
Change millilitres to litres: 100/1000 = 0.1 L
[KOH](dissolved) = [OH-] = 1.996236 x 10-3 mol/0.1 L = 1.996236 x 10-2 M
Kw = [OH-][H+] = 1.0 x 10-14
[H+] = 1.0 x 10-14/[OH-]
= 1.0 x 10-14/1.996236 x 10-2
= 5.0094 x 10-13 M
= 5.0 x 10-13 M
4.
a)
b)
c)
Using Le Chatelier’s principle, what effect will an increase of temperature have on the
following system?
H2O(l) + 57 kJ
H+(aq) + OH-(aq)
What is the expression for the constant for this reaction?
What effect will an increase in temperature have on the value of the constant? Why?
d)
a)
b)
c)
d)
5.
Justify the following statement:
"A decrease in temperature decreases the value
of Kwater"
[H+] and [OH-] increase
K = [H+][OH-]
K will increase because there is more dissociation
K will decrease because there is less dissociation
What is the concentration of H+ in a solution containing 73.0 g of HCl in 0.5 L of water?
1 mol of HCl has a mass of 36.5 g
x mol of HCl has a mass of 76 g
x = 2 mol of HCl in 0.5 L
There is 2 mol of HCl in 0.5 L
There is x mol of HCl in 1 L
x = 4 mol of HCl in 1 L
= 4 M + 1.0 x 10-7 M (Too small to make any difference)
= 4.0 M
6.
What is the [H+] of a solution obtained by adding 0.10 g of KOH to 1.0 L of water?
1 mol of KOH has a mass of 56 g
x mol of KOH has a mass of 0.1 g
x = 0.0018 mol of KOH in 1 L
[KOH] = 0.0018 M = [OH-]added
= 0.0018 M + 1.0 x 10-7 M (Too small to make any difference)
= 0.0018 M
Kw = [OH-][H+] = 1.0 x 10-14
So, [H+] = 1.0 x 10-14 = 1.0 x 10-14 = 5.56 x 10-12 M
[OH-] 0.0018
[H+]
7.
= 5.56 x 10-12 M
= 5.6 x 10-12 M
What are [H+] and [OH-] for 0.01 M perchloric acid?
[HClO4] = 0.01 M = [H+]added
= 0.01 M + 1.0 x 10-7 M (Too small to make any difference)
= 0.01 M
Kw = [OH-][H+] = 1.0 x 10-14
So, [OH-] = 1.0 x 10-14 = 1.0 x 10-14 = 1.0 x 10-12 M
[H+]
0.01
[OH-] = 1.0 x 10-12 M
So, [H+]=10-2 M and [OH-]=10-12 M
8.
Enough water is added to 100 g of sodium hydroxide in order to obtain a one litre solution.
What is the concentration of OH- ions in this solution?
NaOH  Na+ + OH100g/(22.98977 + 15.9994 + 1.0079) = 100/39.99707 = 2.5 mol
M = 2.5 mol/1 L
= 2.5
[NaOH](dissolved) = [OH-] = 2.5 M
```