中4化學知識

Q1. (b) seems to be (100/1000) x 1 = 0.1mol, but it is a weak acid, which only partially ionize; so, the no. of mole of H+ should be less than 0.1(c) (200/1000) x 0.5 = 0.1mol(d) seems to be (100/1000) x 0.5 x 2 = 0.1mol, but some H2CO3 are ionize to hydrogencarbonate ion (HCO3-);so, the no. of mole of H+ should be less than 0.1

Q2.
Concentration = 1.22g cm^3 = 1220g dm^3Molar mass of sulphuric acid = 2(1) + (32.1) + 4(16) = 98.1Molarity = Concentration/molar mass = 1220/98.1Neither one is the answer?!

Q3.
No. of moles of H+ in 25cm^3 1M HCl = (25/1000) x 1 = 0.025mola) No. of moles of H+ in 50cm^3 1M HNO3 = (50/1000) x 1 = 0.05mol No. of moles in total = 0.075 [H+] in total = no. of mole/volume = 0.075/[(25+50)/1000] = 1 pH = -log[H+] = 0b) Ethanoic acid is a weak acid, which partially ionize. (Equilibrium constant of ethanoic acid = 1.8 x 10^-5) Ka = [H+]*[CH3COO-] / [CH3COOH] 1.8 x 10^-5 = (x*x) / 1 x = [H+] = 0.0042426 No. of mole of H+ in 50cm^3 1M CH3COOH = 0.0042426 x (50/1000) = 0.0021213 No. of mole in total = 0.0021213 + 0.025 = 0.046213 [H+] in total = no. of mole/volume = 0.046213/[(25+50)/1000] = 0.61617 pH = -log[H+] = 0.2c) No. of moles of H+ in 50cm^3 1M H2SO4 = (50/1000) x 1 x 2 = 0.1mol No. of mole in total = 0.1 + 0.025 = 0.125 [H+] in total = no. of mole/volume = 0.125/[(25+50)/1000] = 1.6667 pH = -log[H+] = -0.22186

Q4.
No. of moles of H+ = (25/1000) x 0.03 = 0.00075molNo. of moles of OH- = (25/1000) x 0.01 = 0.00025molBy H(+) + OH(-) --------> H2ONo. of mole of H+ left = 0.00075 – 0.00025 = 0.0005mol[H+] after reaction = 0.0005/[(25+25)/1000] = 0.01pH = -log[H+] = 2

Q5. By 2H(+) + CO3(2-) --------> H2O + CO2All carbonates will react with dil. H2SO4 to give CO2a) H2SO4 + Na2CO3 --------> Na2SO4 + H2O + CO2b) H2SO4 + CaCO3 --------> CaSO4 + H2O + CO2c) H2SO4 + MgCO3 --------> MgSO4 + H2O + CO2

2011-08-07 00:45:38 補充：
Q6.
a) 0.2mol H(+) + 0.2mol OH(-)
No. of moles H+ and OH- reacted = 0.2
b) 0.1mol H(+) + 0.2mol OH(-)
No. of moles H+ and OH- reacted = 0.1
c) 0.2mol H(+) + n mol OH(-) which 0.1 < n < 0.2
No. of moles H+ and OH- reacted = n

2011-08-07 00:45:59 補充：
d) m mol H(+) + 0.2mol OH(-) which 0.1 < m < 0.2
No. of moles H+ and OH- reacted = m
As the solutions are of the same volume, and (a) reacted the most, the reaction in (a) gives out most heat.

2011-08-07 00:46:49 補充：
Factors affecting temperature rise
*the volume of solution
*the amount of H(+) and OH(-) reacted
For solution of same volume, more H(+) and OH(-) reacted means more heat released, vice versa.
For solution of same amount of H(+) and OH(-) reacted, smaller volume means more heat released, vice versa.

2011-08-07 00:47:05 補充：
Q7.
No. of moles of Na2CO3 in 250cm^3 solution = mass/molar mass = 1.06/(106) = 0.01mol
No. of moles of Na2CO3 in 25cm^3 solution = 0.01 x (25/250) = 0.001mol
By 2H(+) + CO3(2-) --------> H2O + CO2
No. of moles of HCl required = 0.001*2 = 0.002mol
Volume of HCl = 0.002/0.2 = 0.01dm^3 = 10cm^3

2011-08-07 00:47:44 補充：
Q8.
No. of mole of H+ = 0.1 x (25/1000) x 2 = 0.005mol
By H(+) + OH(-) --> H2O
No. of moles of OH(-) reacted = 0.005mol
No. of moles of XOH = 0.005 x (25/250) = 0.05mol
No. of moles of XOH = mass/molar mass
Molar mass of XOH = mass/no. of moles of XOH
X + 16 + 1 = 2.8/0.05
X = 56 – 16 – 1 = 39

2011-08-07 00:47:58 補充：
Q9.
No. of mole of HCl = 0.03 x (25/1000) = 0.00075mol
No. of mole of Ba(OH)2 = 0.01 x (25/1000) = 0.00025mol
No. of mole of H(+) after reaction = 0.00075 – 0.00025 = 0.0005mol
[H+] = 0.0005/[(25+25)/1000] = 0.01M

2011-08-07 00:48:09 補充：
Q10.
No. of mole of H(+) = 0.05 x (25/1000) x 2 = 0.0025mol
No. of mole of KOH = 0.2 x (25/1000) = 0.005mol
No.of mole of OH(-) after reaction = 0.005 – 0.0025 = 0.0025mol
[OH-] = 0.0025/[(25+25)/1000] = 0.05M

2011-08-07 00:48:30 補充：
Q11.
No. of mole of Na+ = 0.2 x (25/1000) = 0.005mol
[Na+] = 0.005/[(25+25)/1000] = 0.1M

Q12.
No. A base(鹽基) should react with H+ to form H2O and a salt as only products. But Na2CO3 react with H+ to form H2O, salt and CO2. So, it is not a base.

2011-08-07 00:48:53 補充：
Q13.
(i) No. of mole of H+ in acid = 1 x (80/1000) = 0.08mol
No. of mole of OH- in alkali = 1 x (20/1000) = 0.02mol
(ii) No. Because 0.08 mole of acid need 0.08 mole of alkali to neutralize, but there is only 0.02 mole of alkali. So, the neutralization is not complete.

2011-08-07 00:49:01 補充：
(iii) Alkali is the limiting agent(限量反應物).
(iv) By H(+) + OH(-) --------> H2O
No. of mole of water formed = 0.02mol
To complete a neutralization, the no. of mole of H(+) reacted should equal to no. of mole of OH(-) reacted; besides, water and a salt should be the only products.

2011-08-07 00:50:39 補充：
Sorry for using English (I don’t know the terms in Chinese).
Hope I can help you!!!~ ^^
黑白灰_唯物主義者_魑魅魍魎

2011-08-07 23:05:53 補充：
1a)
It is true that ethanoic acid is a weak acid, and it partially ionize.
But, maybe you haven’t learnt, the extent of ionization of a weak acid(and weak alkali) is maintained constant, called equilibrium constant.
(Note that, each weak acid and weak alkali has its own equilibrium constant)

2011-08-07 23:06:10 補充：
1b)
That means, the ratio between [H+]*[CH3COO-] and [CH3COOH] is a constant value. (For ethanoic acid, the equilibrium constant = 1.8 x 10^-5)
In reaction, whenever H+ is used, CH3COOH will ionize to give H+ and CH3COO- to maintain the equilibrium constant.

2011-08-07 23:07:23 補充：
1c)
So, as we don’t know whether the acid or alkali is limiting agent, we use the maximum amount of H+ that the acid can give for calculation.

2011-08-07 23:07:51 補充：
1d)
**We consider the strength only there is no reaction take place.**
**When there is reaction, we do calculation with the ASSUMPTION that the acid/alkali is completely ionized**

2011-08-07 23:08:24 補充：
2a)
By CH3COOH + KOH ---> CH3COOK + H2O (ionic equation :H+ + OH- ---> H2O)
1 mole of CH3COOH reacts with 1 mole of KOH to form 1 mole of H2O
As KOH is limiting agent, we know that only 0.02 mole of H+ in CH3COOH will react

2011-08-07 23:08:45 補充：
2b)
So, as we now have 0.02 mole of CH3COOH reacts with 0.02 mole of KOH, we can know, from the equation, 0.02 mole of water is formed.

2011-08-07 23:08:57 補充：
3)
Question (3)
The pH of 25cm^3 1M HCl = 0
a) After adding HNO3, pH is still 0
b) After adding ethanoic acid, pH is 0.2
c) After adding H2SO4, pH is -0.22186
(Calculation step are shown in my original answer)
So, adding (b) and (c) will lead to a change in pH value.

2011-08-07 23:16:23 補充：
(1a~d) is answering
Q13的 (i)No. of mole of H+ in acid = 1 x (80/1000) = 0.08mol
but as you said it is a weak acid,which only partially ionize,
那乙酸的H+摩爾數應該少於0.08 ??還是某些情況才要考慮強弱??
and
1.8 x 10^-5 怎來的

(2a&b) is answering
生成水的摩爾數為什麼是0.02mol

(3) is answering
question3 agian.

2011-08-07 23:17:22 補充：
If you have any question, you may e-mail me.
My e-mail is [email protected]