Hydrogen ion concentration and the pH scale

Question 1

• strong acids are assumed to be……………completely when dissolved in……………..
• this means the hydrogen ion concentration is related……………………..to the conc. of the acid

Answer 1

• dissociated completely………………………in water

- directly related

Question 2

a solution of HCl of concentration 0.100 mol dm-3 will produce a hydrogen ion concentration of

Answer 2

0.100 mol dm-3

Question 3

the pH of an aqueous solution is related to ……………………………………..through the equation:

Answer 3

-the hydrogen ion concentration

pH = -lg [H+] or pH = lg x 1/[H+]

Question 4

the hydrogen ion concentration, [H+] is measured in

Answer 4

mol dm-3

Question 5

lg =

Answer 5

logarithm which is used to convert a scale of numbers in powers of 10 to a linear scale
e.g.:
lg 100 = 2 (as 100 is 10 to the power of 2)
lg 10 = 1
lg 1 = 0 (as 1 is 10 to the power of 0)
lg 0.01 = -2 (as 0.01 is 10 to the power of -2)

to do this on your calculator, enter log10 (………)
lg is log

Question 6

to calculate the hydrogen ion concentration, use the equation :

Answer 6

[H+(aq)] = 10 to the power of -pH

[H+(aq)] = 10(-pH)

Question 7

for weak acids, there is a significant amount of …………………………acid present in solution

Answer 7

undissociated acid

Question 8

using HA to represent a weak acid, the equation for its dissociation in aqueous solution is:

Answer 8

HA(aq) ⇌ H+(aq) + A-(aq)

Question 9

using the equation for the dissociation of HA and the equilibrium law, we obtain the equation:

Answer 9

[H+(aq)] [A-(aq)]
———————- = a constant (Ka)
[HA(aq)]

Question 10

the acid dissociation constant is given the symbol:

Answer 10

Ka

Question 11

calculate, in steps, the hydrogen ion concentration of an aqueous solution of ethanoic acid of concentration 0.05 mol dm-3, with Ka value 1.74 x10-5 mol dm-3 at 298K.
CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)

Answer 11

1) [CH3COO-(aq)] [H+(aq)]
Ka = ——————————–
[CH3COOH(aq)]
2)
every time a molecule of CH3COOH dissociates, a CH3COO- ion and a H+ ion are formed. this means that [CHCOO-(aq)] = [H+(aq)]
3)
therefore, the expression can be simplified to:
Ka= [H+(aq)] (2) / [CH3COOH(aq)]
4)
Ka = [H+(aq)] (2) / 0.05 = 1.74 x10-5 mol dm-3
[H+(aq)] = (0.5 x 1.74 x10-5) (1/2) = 9.33 x 10-4 mol dm-3

Question 12

when calculating hydrogen ion concentration using the equilibrium law, it is important to note that the concentration used in the expression for Ka are the concentrations at

Answer 12

equilibrium

Question 13

the equation for calculating the hydrogen ion concentration of an aqueous solution of a weak monobasic acid is:

Answer 13

[H+(aq)] = √(Ka x [acid])

where Ka is the dissociation constant for the weak acid

Question 14

pKa =

Answer 14

pKa = -lg Ka

Question 15

• the larger the value of Ka, the…………………..the acid
• the smaller the value of pKa, the ……………………the acid

Answer 15

stronger

stronger

Question 16

some weak monobasic acids include (in increasing strength order):

Answer 16

• propanoic acid
• ethanoic acid
• benzoic acid
• methanoic acid
• chloroethanoic acid
• dichloroethanoic acid
• trichloroethanoic acid

Question 17

the most common dibasic acid is

Answer 17

sulfuric acid

Question 18

sulfuric dissociates in two stages:

Answer 18

H2SO4(aq) –} H+ (aq) + HSO4- (aq)

HSO4-(aq) ⇌ H+(aq) + SO4 2-(aq)

Question 19

H2SO4 is a …………….acid and therefore dissociates…………..
HSO4- is a ………………..acid and therefore dissociates……………….

Answer 19

H2SO4 is a strong acid and therefore dissociates completely

HSO4- is a weak acid and therefore does not dissociate fully

Question 20

the pH of an aqueous solution is defined as

Answer 20

the reciprocal of the logarithm to the base 10 of the hydrogen ion concentration measured in moles per cubic decimeter, pH = -lg [H+].