Neutralization?

Question

Just a few questions.

1. Find the gram equivalent mass for H3PO4

2. Find the gram equivalent mass for Mg(OH)2.

3. How many equivalents are in 10.5 grams of Al(OH)3?

4. How many equivalents are in 5.4 grams of Ca(OH)2?

5. Find the normality of 2.4 M H2CO3.

6. Find the normality of 0.65 M Ba(OH)2 .

7. What is the normality of a solution containing 2.4 g of H3PO4 in 350 mL of solution?

8. Complete the following equation:

Ca(OH)2 + HCl →
9. What volume of 0.58M HF is needed to neutralize 55.2mL of 0.79M Al(OH)3?

10. What is the molarity of Ba(OH)2 if 65.2mL are neutralized by 154.2 mL of 0.50 M HCl?

11. What volume of 0.24 M LiOH is needed to neutralize 125 mL of 0.85 M H2SO4?

12. What volume of 0.050 M Mg(OH)2 is necessary to neutralize 10.5 mL of 0.54 M HNO3?

13. A 25.3 g sample of KOH in 150.5 mL of solution is reacted with 0.75 M H2SO4. What volume of the acid is needed to neutralize the base?

I know there's a lot, but this is a dire emergency. I honestly have no idea what to do. Please, help me. We have a substitute that doesn't explain things in a way that I don't understand, and my parents are clueless. I'd ask others in the class, but I'm homeschooled on the computer.

P.S.- There were 40 questions before these ones on the paper, but I narrowed it down after 3 hours.

Answer 1

1. Gram equivalent mass = molecular weight / 3 = 98/3 = 32.7

2. 58.3 / 2 = 29.15

3. gram equivalent mass = 78 / 3 = 26
equivalents = 10.5 / 26 = 0.404

4. gram equivalent mass = 74 / 2 = 37
5.4 / 37 = 0.146 equivalents

5. N = 2.4 x 2 = 4.8

6. N = 0.65 x 2 = 1.3

7 . 98/3 = 32.7
2.4 g / 32.7 = 0.0734 equivalents
0.0734 / 0.350 = 0.210 N

8. Ca(OH)2 + 2HCl >> CaCl2 + 2H2O

9. Moles Al(OH)3 = 55.2 x 0.79 /1000 = 0.0436
Moles HF needed = 3 x 0.0436 = 0.131
V = moles / M = 0.131/ 0.58 = 0.225 L

10. Moles HCl = 0.50 x 154.2 /1000 = 0.0771
Moles Ba(OH)2 = 0.0771 /2 = 0.0386
M = 0.0386 / 0.0652 = 0.591 M

11. Moles H2SO4 = 125 x 0.85 /1000 = 0.106
moles LiOH needed = 2 x 0.106 = 0.212
V = moles / M = 0.212 / 0.24 = 0.885 L

12. Moles HNO3 = 10.5 x 0.54 /1000 = 0.00567
moles Mg(OH)2 needed = 0.00567 / 2 = 0.00283
V = 0.00283 / 0.050 = 0.0567 L

13. Molecular weight KOH = 56 g/mol
25.3 / 56 = 0.452 mole KOH
Mole H2SO4 needed = 0.452 /2 = 0.226
V = 0.226 / 0.75 = 0.301 L

Answer 2

While doing titration experiments this a very common problem. However you need to be patient in these experiments, mixing a drop a time from the time you see the pink color for the first time. Make sure that you continuosly stir the beaker. Placing a white paper sheet below the beaker always helps. A temporary color is not the right indication of the neutral point. If the color disappears on stirring the solution, it is not the neutral point.

Answer 3

1. Mol. wt. H3PO4 = 98; 98/3 = 32.7g/eq

2. Mol. wt. Mg(OH)2 = 58; 58/2 = 29g/eq

3. Mol. wt. Al(OH)3 = 78; 78/3 = 26g/eq

4. Mol. wt. Ca(OH)2 = 74; 74/2 = 37g/eq

5. 2.4M H2CO3 = 4.8N

6. 0.65M Ba(OH)2 = 1.30N

7. Let solution be called S. Mol. wt. H3PO4 = 98

2.4gH3PO4/350mLS x 1molH3PO4/98gH3PO4 x 2eqH3PO4/1molH3PO4 x 1000mLS/1LS = 0.14eqH3PO4/1LS = 0.14N

8. Ca(OH)2 + 2HCl ===> CaCl2 + 2H2O