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A2 Chem Help: Acids and Bases Question
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I'm on a-levelchemistry.co.uk at the moment doing Unit 4 Acid and Base questions.
How do you know if it is an acid or base?
Bronsted lowry acid- H+ donor
Bronsted lowry base- H+ acceptor
1. State whether the following species can behave as an acid, as a base, or both:
a) NH4+ (A: Base)
b) NH3 (A: Acid)
c) H3O+ (A: Acid and Base)
d) HClO4 (A: Acid)
e) CO32- (A: Acid and Base)
f) NO3- (A: Base)
g) CH3CH2OH (A: Acid and Base)
h) CH3COOH (A: Acid and Base)
i) HSO4- (A: Acid and Base)
j) HNO3 (A: Acid and Base)
k) HCl (A: Acid and Base)
This is meant to be an easy question. Thanks!
How do you know if it is an acid or base?
Bronsted lowry acid- H+ donor
Bronsted lowry base- H+ acceptor
1. State whether the following species can behave as an acid, as a base, or both:
a) NH4+ (A: Base)
b) NH3 (A: Acid)
c) H3O+ (A: Acid and Base)
d) HClO4 (A: Acid)
e) CO32- (A: Acid and Base)
f) NO3- (A: Base)
g) CH3CH2OH (A: Acid and Base)
h) CH3COOH (A: Acid and Base)
i) HSO4- (A: Acid and Base)
j) HNO3 (A: Acid and Base)
k) HCl (A: Acid and Base)
This is meant to be an easy question. Thanks!
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(Original post by Pigster)
a) A
b) B
c) A
d) A
e) B
f) B
g) both
h) both
i) both
j) both
k) A
I think.
a) A
b) B
c) A
d) A
e) B
f) B
g) both
h) both
i) both
j) both
k) A
I think.
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(Original post by Pigster)
a) A
b) B
c) A
d) A
e) B
f) B
g) both
h) both
i) both
j) both
k) A
I think.
a) A
b) B
c) A
d) A
e) B
f) B
g) both
h) both
i) both
j) both
k) A
I think.
Also...
Do you know how you would work out this question?
Calculate the pH of 0.002 mol dm-3 of KOH using the equation pH= -lg[H+]
The answer is 11.3 but I got 2.70
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#5
(Original post by i'mlaura)
Also...
Do you know how you would work out this question?
Calculate the pH of 0.002 mol dm-3 of KOH using the equation pH= -lg[H+]
The answer is 11.3 but I got 2.70
Also...
Do you know how you would work out this question?
Calculate the pH of 0.002 mol dm-3 of KOH using the equation pH= -lg[H+]
The answer is 11.3 but I got 2.70
you need Kw to work out this question
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(Original post by Farseer)
KOH is an alkali it gives out OH- not H+
you need Kw to work out this question
KOH is an alkali it gives out OH- not H+
you need Kw to work out this question
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#7
(Original post by i'mlaura)
Thank you, that's correct for most of them. How did you know that though?
Thank you, that's correct for most of them. How did you know that though?
a) NH4+ (A: Base) NH4+ can donate an H+ so acts as an acid
b) NH3 (A: Acid) NH3 has a lone pair, so can accept - base
c) H3O+ (A: Acid and Base) H3O+ has a lone pair, so could accept another H+, but is nearly impossible
d) HClO4 (A: Acid) H2ClO4+ can exist, but wouldn't like it either
e) CO32- (A: Acid and Base) CO32- doesn't contain H, so can't donate H+, so can't be an acid
f) NO3- (A: Base) correct
g) CH3CH2OH (A: Acid and Base) ethanol can accept protons, but isn't especially happy about it. Does the reaction between ethanol and Na involve H+ donation? is it an acid?
h) CH3COOH (A: Acid and Base) ethanoic acid is deffo an acid, but can be protonated by stronger acids
i) HSO4- (A: Acid and Base) correct
j) HNO3 (A: Acid and Base) nitric is usually an acid, but can be protonated by stronger acids e.g. nitration of benzene
k) HCl (A: Acid and Base) hydrochloric is deffo an acid, H2Cl+ can form, but like H4O2+ isn't too happy about it (OK slightly less grumpy)
You don't have to use Kw to work out the pH of strong bases:
pOH = -log[OH-] = -log 0.002 = 2.70
pH = 14.00 - pOH = 14.00 - 2.70 = 11.30
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#8
(Original post by Pigster)
You don't have to use Kw to work out the pH of strong bases:
pOH = -log[OH-] = -log 0.002 = 2.70
pH = 14.00 - pOH = 14.00 - 2.70 = 11.30
You don't have to use Kw to work out the pH of strong bases:
pOH = -log[OH-] = -log 0.002 = 2.70
pH = 14.00 - pOH = 14.00 - 2.70 = 11.30
pH + pOH = pKw
and 14 that you used to convert pOH to pH is nothing else but pKw.
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