SOLVED: ' A solution of sulfuric acid (H2 SO4, 25.00 mL) was titrated to the second equivalence point (both protons were) with 34.55 mL of 0.1020 M sodium hydroxide. What was the concentration of the sulfuric acid? a. 0.07048 M b. 0.1410 M c. 0.2819 M d 0.0353 M e 0.0533 M What volume of 0.80 M HCI will be required to titrate a 20.0 mL solution of 0.60M NaOH to the equivalence point? a. 15 mL b. 20 mL c.25 mL d. 30 mL e. None of the above What volume of 0.80 M HCI will be required 36.2 grams of NaBrO (fw = 118.9 g/mol) to the equivalence point? a. 150 mL b. 381 mL c.b. 258 mL d. 308 mL e None of the above One brand of extra

您所在的位置：网站首页 › maven could not calculate › SOLVED: ' A solution of sulfuric acid (H2 SO4, 25.00 mL) was titrated to the second equivalence point (both protons were) with 34.55 mL of 0.1020 M sodium hydroxide. What was the concentration of the sulfuric acid? a. 0.07048 M b. 0.1410 M c. 0.2819 M d 0.0353 M e 0.0533 M What volume of 0.80 M HCI will be required to titrate a 20.0 mL solution of 0.60M NaOH to the equivalence point? a. 15 mL b. 20 mL c.25 mL d. 30 mL e. None of the above What volume of 0.80 M HCI will be required 36.2 grams of NaBrO (fw = 118.9 g/mol) to the equivalence point? a. 150 mL b. 381 mL c.b. 258 mL d. 308 mL e None of the above One brand of extra

SOLVED: ' A solution of sulfuric acid (H2 SO4, 25.00 mL) was titrated to the second equivalence point (both protons were) with 34.55 mL of 0.1020 M sodium hydroxide. What was the concentration of the sulfuric acid? a. 0.07048 M b. 0.1410 M c. 0.2819 M d 0.0353 M e 0.0533 M What volume of 0.80 M HCI will be required to titrate a 20.0 mL solution of 0.60M NaOH to the equivalence point? a. 15 mL b. 20 mL c.25 mL d. 30 mL e. None of the above What volume of 0.80 M HCI will be required 36.2 grams of NaBrO (fw = 118.9 g/mol) to the equivalence point? a. 150 mL b. 381 mL c.b. 258 mL d. 308 mL e None of the above One brand of extra

2023-03-24 16:51| 来源: 网络整理| 查看: 265

Transcript

The first thing we need to do is write down the chemical reactions, which include hydrochloric acid plus sodium hydroxide. This is already balanced, so we could just say that we have 10 liters of hydrochloric acid. The number of moles here would be equivalent to 2.010 liters. We have.010 moles of hydrochloric acid because this would be multiplied by 1.0 molars. Adding sodium hydroxide for part is what we're doing. The mole ratio is 1 to 1 because we have 2 milliliters at 1 molar concentration and so we can actually multiply these 2 and get 20 moles of sodium hydroxide. This is for these. We can say that the concentration of h of hydrogen ion or in this case protons, would be equal to 2.010 moles of hydrochloric acid, minus.0020 moles of sodium hydroxide, and this would all be divided by 2 liters. The ph would be equal to the negative log base 10 of the hydrogen ion concentration, and this is very different at x, because this gives us a concentration of 1.667 molar. For part b, we have both hydrochloric acid and sodium hydroxide, but they are both 10 millimeters at 1.0 molar concentration. Again, they have a 1 to 1 mole ratio as deduced by the chemical reaction, and so we could say that this would be perfectly balanced or we have a ph equal to 7 point. This will result in salt water. It's better to figure out the concentration of hydroxide ion than it is to find out what it's acidic for. This would be the same amount of sodium hydroxide as................. At 1.0 molar, this would be minus the 10 liter.010 liters, and then this would be divided by.012, plus.012 rot, and so we have a concentration of.020. We can say that the p o h is equal to the negative logarithm base 10 again. This is giving us 1.3 approximately, and so we know that p h, plus p, o h would be equal to 14 point, and so we could say the p h would be equaling to approximately 12.7 or 14 minus 1.3 12.7. We have some hydrochloric acid again. We have 50 millimeters or.050 liters at a concentration. This has given us our.025 molar, and so we are reacting hydrochloric acid with sodium carbonate. Let's first write our chemical reaction, if we are reacting. So we have 2 h, c, l plus and a 2 c, o 3 of the sodium carbonate, and this is going to react to form 2 n, a c l plus igh 2, o plus carbon dioxide. The ratio of…

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