F325 Acids, Bases and Buffers Hw

1.	Mass of HNO3 = 
Moles of HNO3 = 
pH = –log[H+] = –log 14.4 = –1.16/1.2  calc –1.15836
pH from ignoring 65% pH = –1.35: with working, 2 marks.
[3]
 
2.	[H+] increases (1)
H2O ionises more /
for H2O  H+ + OH–, equilibrium moves to the right (1)	2
exo/endo is ‘noise’
[2]

3.	(i)	proton donor (1)	1
(ii)	partially dissociates (1)	1
[2]

4.	C6H5OH(aq) + OH–(aq)  C6H5O–(aq) + H2O(l)
	acid 1	base 2	base 1	acid 2 (1)
[1]

5.	(i)	Ka = [C6H5O–(aq)] [H+(aq)] / [C6H5OH(aq)] (1)	1
(ii)	Mr C6H5OH = 94 (1)
[C6H5OH(aq)] 4.7/94 = 0.050 mol dm–3 (1)
1.3 × 10–10 ≈ [H+(aq)]2 / 0.050 mol dm–3 (1) (‘=’ sign is acceptable)
[H+] = √{(1.3 × 10–10) × (0.050) } = 2.55 × 10–6 mol dm–3 (1)
pH = –log[H+] = –log 2.55 × 10–6 = 5.59 (1)
	3 marks: [H+]; pH expression ; calc of pH from [H+]	5
[6]

6.	proton donor 
partially dissociates 
[2]

7.	Ka = 
1.58 × 10–4 = / 
[H+] = √{(1.58 × 10–4) × (0.025) } = 1.99 × 10–3 mol dm−3 
pH = −log[H+] = −log 1.99 × 10–3 = 2.70 
5.4034 (no square root) with working would score 1 mark.
[3]
8.	(i)	Kw = [H+(aq)] [OH–(aq)] (1)	1
state symbols not needed
(ii)	[H+(aq)] = 10–pH = 10–13.54 = 2.88/2.9 × 10–14 mol dm–3 (1)
[NaOH] / [OH–(aq)] = 
= 0.347 / 0.35 mol dm–3 (1)	2
[3]

9.	(a)	partial dissociation: HCOOH  H+ + HCOO– (1)	1
(b)	(i)	pH = –log (1.55 × 10–3) = 2.81/2.8 (1)
[H+] deals with negative indices over a very wide range/
pH makes numbers manageable
/removes very small numbers (1)	2
(ii)	Ka =  (1) (state symbols not needed)	1
(iii)	Ka =  (1)
= 1.60 × 10–4 (mol dm–3)(1)
pKa = –log Ka = –log (1.60 × 10–4) = 3.80 (1)	3
(iv)	Percentage dissociating =  = 10.3 % /
10% (1)	1
(working not required)
[8]

10.	(a)	strength of acid/extent of dissociation/ionisation (1)	1
(b)	(i)	H2SO3(aq) + CH3COOH(aq)  HSO3–(aq) + CH3COOH2+(aq)
acid 1	base 2 (1)	base 1	acid 2 (1)	2
1 mark for labels on each side of equation
(ii)	CH3COOH is the stronger acid/
Ka CH3COOH is greater/
CH3COOH is more acidic ORA (1)
	C6H5OH(aq) + CH3COOH(aq)  C6H5OH2+(aq) + CH3COO–(aq) (1)	2
 (c)	For HCl, pH = –log[H+] (1) (or with values).
Could be awarded below
= –log 0.045 = 1.35 (1) (accept 1.3)
	For CH3COOH, [H+] = √(Ka × [CH3COOH]) /
√(1.70 × 10–5 × 0.045) (1)
[H+] = 8.75 × 10–4 mol dm–3 (1)
pH = –log 8.75 × 10–4 = 3.058/3.06 (1) (accept 3.1)	5
[10]

11.	Two species differing by H+ …. AW
	one pair:    HNO3 and NO3– 
other pair:  HCOOH and HCOOH2+ 
[3]

12.	(i)	I2(aq) + H2S(g) → 2HI(aq) + S(s)
species and balance (1)
state symbols: accept (s) for I2; (aq) for H2S (1)	2
 (ii)	amount I2 reacted = 1.89 mol / HI formed = 3.44 mol (1)
theoretical amount HI produced = 3.78 mol/484 g (1)
% yield =  = 91.0 % (1)	3
 (iii)	[HI] =  = 4.58/4.59 mol dm–3 (1)
pH = –log 4.59 = –0.66 (1)	2
[7]

13.	(i)	A solution that minimises pH changes/resists pH
changes/opposes pH changes 
(not pH is kept constant/pH maintained/pH cancelled out.	1
(ii)	HCOONa/HCOO–/ NaOH 
HCOO– is the conjugate base/
HCOONa is the salt of the weak acid or HCOOH/
HCOONa supplies HCOO– 	2
 (iii)	Two points from:
Ka /pKa /acid strength/amount of dissociation 
temperature  (but not “temperature & pressure”)
ratio/amounts/concentrations of weak acid and
conjugate base/salt  (or reverse ratio)
(not ………. concentration of base as it could imply
NaOH)	2 max
[5]

14.	

/ NaOH /Na (1)
weak acid/base pair mixture formed (1)	2
On structure, 1 mark for O Na on either or both phenol groups.
[2]
 15.	→ ………… CO2 + H2O 
	Complete correct balanced equation for 2nd mark:
2HNO3 + CaCO3 → Ca(NO3)2 + CO2 + H2O /
2H+ + CaCO3 → Ca2+ + CO2 + H2O /
2H+ + CO32– → CO2 + H2O 
[2]

16.	HCl and CH3COOH have same number of moles/
release same number of moles H+ /
1 mole of each acid produce ½ mol of H2 (1)
	[H+] in CH3COOH < [H+] in HCl/
CH3COOH is a weaker acid than HCl (ora) (1)
	Mg + 2HCl → MgCl2 + H2 (1)
Mg + 2CH3COOH → (CH3COO)2Mg + H2 (1)
	or
Mg + 2H+ → Mg2+ + H2 (1)(1)	4
[4]
 
17.	(a)	Ca3(PO4)2 + 2H2SO4 → Ca(H2PO4)2 + 2CaSO4 (1)	1
(b)	H2PO4–(aq)  H+(aq) + HPO42–(aq) /
	H2PO4–(aq)  2H+(aq) + PO43–(aq) (1)
	(or equivalent with H2O forming H3O+)	1
(c)	(i)	HPO42– (1)	1
(ii)	H3PO4 (1)	1
(iii)	H2PO4– produced Ca(H2PO4)2 or on LHS of an attempted
equilibrium equation (1)
2 equations/equilibria to shown action of buffer (1)(1)
from:
H2PO4– + H+  H3PO4 /
H2PO4–  H+ + HPO42–/
H2PO4– + OH–  H2O + HPO42– /
H+ + OH–  H2O	3
[7]

18.	amount of NaOH in titration = 0.175 x 22.05/1000
or 3.86 × 10–3 (1) (calc: 3.85875 x 10–3)
	amount of A in 25.0 cm3 = 0.5 × mol NaOH
or 1.93 × 10–3 (1) (calc: 1.929375 × 10–3)
	amount of A in 250 cm3 = 10 × 1.93 × 10–3 or 1.93 × 10–2 (1)
	1.93 × 10–2 mol A has a mass of 2.82 g
	molar mass of A = 2.82/1.93 × 10–2 = 146 g mol–1 (1)
(or Mr of A is 146)
	Therefore A is adipic acid / HOOC(CH2)4COOH (1)	5
[5]
 
19.	moles of NaOH =  / 3.00 × 10–3 mol 
moles of acid = 3.00 × 10–3 mol 
moles of acid in flask = 10 × 3.00 × 10–3 = 3.00 × 10–2 mol 
molar mass of compound = 	4
 	Molecular formula = C4H6O2 
	A 4 carbon carboxylic acid
(e.g. butanoic acid) shown (bod) 
	Any 2 possible isomers  from:
CH2 = C(CH3)COOH
CH2 = CHCH2COOH
cis CH3CH = CHCOOH
trans CH3CH = CHCOOH	4
Accept structural formulae that are unambiguous.
[8]

20.	(a)	(i)	(+)1 (1)	1
(ii)

Look for atoms bonded together.
AND other lone pairs.	1
(b)	(i)	C13H18O2 (1)
(ii)	any chemical that reacts to produce gas:
e.g. carbonate and CO2 (1)
accept: metal more reactive than Pb and H2
	balanced equation to match chemical added (1)	3
 (c)	Mr(Lidocaine) = 236 (1)
Moles Novocaine = 100 × 10-3/236 = 4.24 × 10–4 (1)
Concentration of Novocaine = 4.24 × 10–4 × (1000/5)
= 0.0847/0.0848/0.085 mol dm–3 (1)	3
 (d)	mass C = 12 ×  = 1.02 g /
moles CO2 =  = 0.085 mol (1)
mass H =  × 0.918 = 0.102 g /
moles H2O =  = 0.051 mol (1)	2
ratio C : H =  = 0.0850 : 0.102 = 5 : 6 / 10 : 12/
ratio CO2 : H2O = 5 : 3 / 10 : 6 (1)
mass O = 1.394 – (1.020 + 0.102) = 0.272 g
/ using 1.394 g eugenol and Mr = 164, shows that 1
molecule contains 2 atoms of O (1)	2
 molecular formula = C10H12O2 (1)	1
[13]