CHEM.化學 MC

Kyy010203

2010-10-03 11:41 pm

如果加埋Calculation steps就最好!!!!!

Sulphuric acid is added continuously to some potassium hydroxide in a conical flask.
Which of the following is taking place in the beaker?
A. The pH of the solution decreases.
B. The pH of the solution increases.
C. The concentration of the hydrogen ions decreases.
D. Potassium sulphate crystallizes in the beaker.

If 15 cm3 of 0.1M hydrochloric acid is required to neutralize a beaker of sodium
hydroxide solution, we can deduce from the above information
(1) the mass of sodium hydroxide in the solution.
(2) the concentration of sodium hydroxide.
(3) the number of moles of sodium hydroxide in the solution.
A. (1) and (2) only
B. (1) and (3) only
C. (2) and (3) only
D. All of them

An aqueous solution contains 14 g/dm3 of sulphuric acid. The volume of this solution
which would react with 25 cm3 of 0.1M solution of sodium hydroxide is
A. 8.75 cm3
B. 17.5 cm3
C. 35.5 cm3
D. 49.0 cm3
[Formula mass of sulphuric acid = 98]

What is the volume of 0.2M sodium carbonate solution required to react
completely with a solution containing 0.005 mole of sulphuric acid and 0.005 mole of hydrochloric acid?
A. 37.5 cm3
B. 75 cm3
C. 100 cm3
D. 150 cm3

0.795g of pure sodium carbonate is dissolved in distilled water and then made up
to 250 cm3. 25 cm3 of this solution require 15 cm3 of a solution of hydrochloric acid
for complete reaction. The concentration of the hydrochloric acid is
A. 0.05M
B. 0.10M
C. 0.15M
D. 0.20M

回答 (1)

己式庚辛

2010-10-04 7:03 am

✔ 最佳答案

1. A
when sulphuric acid is added, hydroxide is consumed and more and more hydrogen ions are added. pH decreases as it becomes more acidic. thus B and C are incorrect (B&C are just same thing).
D is inappropiate as concentration of solutions is not specified.
(btw, why the container of KOH is not consistent? beaker or conical flask?)

2. D
HCl + NaOH ------> NaCl + H2O
as you have concentration and volume of HCl, you know the no. of mole of HCl:
no. of mole = molarity x volume

as you can write reaction equation, mole ratio and no. of mole of NaOH can also be found.

if you also know volume of NaOH, you can calculate its concentration:
molarity = no. of mole / volume

and, mass = no. of mole x molar mass.

3. B
no. of mole of H(+) given by HCl = 0.005mole
no. of mole of H(+) given by H2SO4 = 0.005x2 = 0.010mole
total no. of mole of H(+) = 0.015mole
no. of mole of OH(-) needed = 0.015mole
volume of OH(-) needed = no. of mole / molarity = 0.015 / 0.2 = 0.075 dm^3
= 75ml

4. B
mass of Na2CO3 in 25cm^3 = 0.0795g
molar mass of Na2CO3 = 106g/mole
no. of mole of Na2CO3 = 0.0795/106 = 7.51E-4 mole
2HCl + Na2CO3 ------> 2NaCl + H2O + CO2
mole ratio of HCl : Na2CO3 = 2:1
no. of mole of HCl = 0.00152 mole
molarity = no. of mole / volume = 0.00152/(15/1000)
= 0.10M

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