Molarity

The answer is as follows :


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2014-08-12 01:32:08 補充：
The solution of each salt is prepared by neutralization of the corresponding acid and the corresponding alkali. As the acid used is a strong acid, methyl orange is chosen as the indicator of neutralization.

2014-08-12 01:32:19 補充：
The indicator shows the end point of the neutralization. At the end point, the acid and the alkali are just completely reacted without any one of them in excess.

thank you so much pingshek!!!

(1) 20.0 cm3 of 1.0 M NaCl(aq) are mixed with 10.0 cm3 of 2.0 M Na2CO3(aq). What's the concentration of Na+(aq)ions in the resulting solution?
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This question requires you to give the concentration not molarity of the solution.
You should answer in gdm-3 instead of M

The number of moles of Na ion in NaCl= 0.02 mol
The number of moles of Na ion in Na2CO3= 0.01*2*2= 0.04 mol

Mass of Na ion in NaCl= 0.02*(23)= 0.46g
Mass of Na ion in Na2CO3= 0.04*(23)= 0.92g

Concentration of Na ion= (0.46+0.92)/0.03 = 46 g dm-3
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(2) Calculate the molarity of the following situations:

10.0 g of KOH in 250.0 cm3 of solution
Concentration= 10/0.25= 40 g/dm3
Molarity= 40/(39.1+16+1)= 0.713 M

7.93 g of (NH)4SO4 in 750.0 cm3 of solution
Concentration= 7.93/0.75= 10.6g/dm3
Molarity= 10.6/156.1=0.0679M

57.1 g of FeCl2 in 2.50 dm3 of solution
Note that 1dm3=1000cm3
Concentration= 57.1/2.5=22.84g/dm3
Molarity= 22.84/90= 0.254M

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(3) A 10.0 cm3 of a sample of domestic sewage has an effluent pH of 6.80.After this sewage has been treated, the effluent pH is 7.00. What's the change in the conc. of H+ ions?

Original H ion concentration= 10^(-6.8)
New H ion concentration= 10^(-7)
Change= 10^(-6.8)-10^(-7) =5.85*10^-8
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2014-08-11 21:11:11 補充：
To determine the end point of the titration at which just enough acid has been added to neutralize the alkali. Otherwise, the resultant salt solution and hence the salt would be contaminated with the excess acid or alkali solution.

2014-08-11 21:17:45 補充：
The molarity of 2c. should be 0.180M. Sorry.