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luluwa_999 luluwa_999
wrote...
Posts: 11
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11 years ago
I want some help in these Question :

1- Calcium nitrate tetrahydrate dissolves in cold water to the extent of 266 g per 100. cm3.
What is the concentration of nitrate ions in this solutio n?
2- 34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an H2SO4 solution.
What is the concentration of the original sulfuric acid solution?
3- 35.0 mL of 0.255 M nitric acid is added to 45.0 mL of 0.328 M Mg(NO3)2. What is the
concentration of nitrate ion in the final solution?
4- What mass of K2CO3 is needed to prepare 200. mL of a solution having a potassium ion
concentration of 0.150 M?
5- What is the molar concentration of chloride ions in a solution prepared by mixing 100.
mL of 2.0 M KCl with 50. mL of a 1.5 M CaCl2 solution?
Thank you
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wrote...
11 years ago
1- Calcium nitrate tetrahydrate dissolves in cold water to the extent of 266 g per 100. cm3.
What is the concentration of nitrate ions in this solutio n?

moles Ca(NO3)2 * 4 H2O = 266 g / 236.15 g/mol=1.13

moles NO3- = 2 x 1.13 = 2.26

concentration NO3- = 2.26 / 0.100 L = 22.6 M
wrote...
11 years ago
2- 34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an H2SO4 solution.
What is the concentration of the original sulfuric acid solution?

2NaOH + H2SO4 ----> Na2SO4 + 2H2O
number of mole of NaOH=(0.1510 X 34.62)/1000=5.23X10^ -3mol
number of mole of H2SO4=(5.23X10^ -3)/2=2.61X10 ^-3mol
original volume of H2SO4=50ml=0.05L
original concentration of H2SO4=2.61X10 ^-3mol/0.05L=0.0522mol/L
wrote...
Educator
11 years ago
3- 35.0 mL of 0.255 M nitric acid is added to 45.0 mL of 0.328 M Mg(NO3)2. What is the
concentration of nitrate ion in the final solution?

Multiply .255 by .0350 L to get .00893 moles HNO3 (sig figs)

Multiply .328 by .0450 L to get .0148 moles Mg(NO3)2 (sig figs). However, there are .0296 moles of nitrate ions.

.00893 mol + .0296 mol = .0385 mol (sig figs)
.0350 L + .0450 L = .0800 L

.0385 mol / .0800 L = .481 M
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