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acid base neutralization problems
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Question:During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200. mL of the base was required to neutralize the sample. How many grams of HCl are present in 500. mL of this acid solution?
Answers:First you need to convert the mL of NaOH to liters Then mulitply by the molarity conversion factor to get moles of NaOH Then use the molemole factor from the reaction equation to convert moles of NaOH to moles of HCl Once you have moles of HCl, convert 50mL to L. Divide the moles of HCl by the L of HCl to find molarity (remember molarity is moles per liter) Here is what it would look like! 200.ml NaOH x (1 L/1000mL) x (0.500 mol NaOH/1 L) x (1 mol HCl/1 mol NaOH) = 0.100 moles HCl 50.0ml x (1 L/1000mL) = 0.050 L 0.100 moles/0.050 L = 2.00 M HCl (3 sig figs!)
Answers:First you need to convert the mL of NaOH to liters Then mulitply by the molarity conversion factor to get moles of NaOH Then use the molemole factor from the reaction equation to convert moles of NaOH to moles of HCl Once you have moles of HCl, convert 50mL to L. Divide the moles of HCl by the L of HCl to find molarity (remember molarity is moles per liter) Here is what it would look like! 200.ml NaOH x (1 L/1000mL) x (0.500 mol NaOH/1 L) x (1 mol HCl/1 mol NaOH) = 0.100 moles HCl 50.0ml x (1 L/1000mL) = 0.050 L 0.100 moles/0.050 L = 2.00 M HCl (3 sig figs!)
Question:I am in AP Chemistry and we have recently started the chapter on acids and bases. I am kind of confused because we just started two days ago, and he is already giving us questions that are ridiculously hard. I could use a little explanation. Anyone?
III. Solve these neutralization equations as indicated (reminder  neutralization occurs when the moles of H+ exactly equal the moles of OH).
1. 25.3 ml of an acid solution whose concentration is unknown is neutralized with 35.8 ml of a base solution whose [OH] is .525 M. Determine the [H+] of the acid solution.
2. 3.45 ml of a 3.45 M HCl solution are neutralized with 8.21 ml of an unknown base. Determine the [OH] for the base.
Again, thanks to anyone that can help!
Answers:Okay. You know that the # of moles OH has to equal the # of moles H+. It's given as the definition of neutralization. 1. The molarity of the base is .525 M. This means you have .525 Moles of OH per liter of solution. If you only have 35.8mL of the solution (the amount needed to neutralize your mystery acid), then you have this many moles of OH: (.525MolesOH/1L solution) * (1L solution / 1000mL solution) = X / 35.8 mL solution Solve for X: X = 35.8mL Solution * (.000525 Moles OH / 1mL solution X = .0188 Moles OH Since you have to have the same number of Moles of OH to neutralize the H+, this means you have .0188 Moles of H+ in your 25.3 mL of mystery acid. Now, work it forward to find the molarity: .0188 moles H+ / 25.3 mL Solution = Y / 1000mL solution Y = 1000mL solution * (.0188 moles H+ / 25.3 mL Solution) Y = .743 Moles H+ / 1000mL Since 1000 mL = 1 L you have .743 Moles H+ per liter, so.... [H+] of your mystery acid = .743M 2. Same theory here. If you have a 3.45M HCl solution, you have a 3.45M H+ solution (assuming 100% dissociation, which is a good assumption here). You find out how many moles of H+ are in 3.45 mL of this solution by: (3.45 Moles H+ /1 Liter) * (1 Liter / 1000mL) = Y / 3.45 mL Y = (3.45 mL) * (3.45 Moles H+ / 1000mL) Y = .0119 moles H+ Now that you know the 8.21 mL of mystery base has to have .0119 moles of OH to neutralize the H+.... .0119 moles OH / 8.21 mL solution = X / 1000 mL solution X = 1000 mL * (.0119 moles OH / 8.21 mL solution) X = 1.45 moles OH in 1000mL of solution. Therefore... [OH] = 1.45M Go forth and burninate many villagers, young Trogdor!
Answers:Okay. You know that the # of moles OH has to equal the # of moles H+. It's given as the definition of neutralization. 1. The molarity of the base is .525 M. This means you have .525 Moles of OH per liter of solution. If you only have 35.8mL of the solution (the amount needed to neutralize your mystery acid), then you have this many moles of OH: (.525MolesOH/1L solution) * (1L solution / 1000mL solution) = X / 35.8 mL solution Solve for X: X = 35.8mL Solution * (.000525 Moles OH / 1mL solution X = .0188 Moles OH Since you have to have the same number of Moles of OH to neutralize the H+, this means you have .0188 Moles of H+ in your 25.3 mL of mystery acid. Now, work it forward to find the molarity: .0188 moles H+ / 25.3 mL Solution = Y / 1000mL solution Y = 1000mL solution * (.0188 moles H+ / 25.3 mL Solution) Y = .743 Moles H+ / 1000mL Since 1000 mL = 1 L you have .743 Moles H+ per liter, so.... [H+] of your mystery acid = .743M 2. Same theory here. If you have a 3.45M HCl solution, you have a 3.45M H+ solution (assuming 100% dissociation, which is a good assumption here). You find out how many moles of H+ are in 3.45 mL of this solution by: (3.45 Moles H+ /1 Liter) * (1 Liter / 1000mL) = Y / 3.45 mL Y = (3.45 mL) * (3.45 Moles H+ / 1000mL) Y = .0119 moles H+ Now that you know the 8.21 mL of mystery base has to have .0119 moles of OH to neutralize the H+.... .0119 moles OH / 8.21 mL solution = X / 1000 mL solution X = 1000 mL * (.0119 moles OH / 8.21 mL solution) X = 1.45 moles OH in 1000mL of solution. Therefore... [OH] = 1.45M Go forth and burninate many villagers, young Trogdor!
Question:In the following acidbase neutralization, 1.68 g of the solid acid HC6H5O (94.12g/mol) neutralized 11.61 mL of aqueous solution base by the reaction:
NaOH + HC6H5O > H2O + NaC6H5O
Calculate the molarity of the base solution.
Answers:Ok, so we need to find out how much NaOH was consumed in the reaction and, thus, how much of it was in the aqueous solution. Taking into consideration the molar masses of NaOH and HC6H5O and the mass of HC6H5O, we get this equation: M naoh / m naoh (consumed) = M hc6h5o / m hc6h50 (consumed), that is 40 / m NaOh = 94.12 / 1.68. Obviously, m NaOH = 0.71 g. To find out the molarity, we need the number of moles, not the mass, so we will convert the mass in moles, by dividing it with the molar mass number: 0.71/40=0.017 moles of NaOH. We will also need the volume in litres, so we have to convert the mL to L, thus having 11.61/1000=0.011 L of aqueous solution. molarity = number of moles / volume = 0.017/0.011 = 1.54 Hope this was clear enough.. :)
Answers:Ok, so we need to find out how much NaOH was consumed in the reaction and, thus, how much of it was in the aqueous solution. Taking into consideration the molar masses of NaOH and HC6H5O and the mass of HC6H5O, we get this equation: M naoh / m naoh (consumed) = M hc6h5o / m hc6h50 (consumed), that is 40 / m NaOh = 94.12 / 1.68. Obviously, m NaOH = 0.71 g. To find out the molarity, we need the number of moles, not the mass, so we will convert the mass in moles, by dividing it with the molar mass number: 0.71/40=0.017 moles of NaOH. We will also need the volume in litres, so we have to convert the mL to L, thus having 11.61/1000=0.011 L of aqueous solution. molarity = number of moles / volume = 0.017/0.011 = 1.54 Hope this was clear enough.. :)
Question:Calculate the molar concentration of 27.0 mL of iron (III) hydroxide needed to neutralize 43.0 mL of 9.3 sulfuric acid. my bad, 9.3 M sulfuric acid Also, could you please show work? THanks
Answers:Fe(OH)3 (s) + 3H2SO4 (aq) => Fe2(SO4)3 (aq) + 3H2O (l) it's given 43.0 mL (or 0.043 L) of H2SO4 9.3M that means the mol of H2SO4 = 0.043 x 9.3 = 0.3999 = 0.4 (mol) the ratio between Fe(OH)3 and H2SO4 is 1 : 3 so if you need to neutralize 0.4 mol H2SO4, u must have 0.4 / 3 = 2/15 mol of Fe(OH)3 the solution with the volume of 27.0 mL or 0.027 L, which contains 2/15 mol Fe(OH)3 will have the molar concentration of 2/15 / 0.027 = 4.938 (M)
Answers:Fe(OH)3 (s) + 3H2SO4 (aq) => Fe2(SO4)3 (aq) + 3H2O (l) it's given 43.0 mL (or 0.043 L) of H2SO4 9.3M that means the mol of H2SO4 = 0.043 x 9.3 = 0.3999 = 0.4 (mol) the ratio between Fe(OH)3 and H2SO4 is 1 : 3 so if you need to neutralize 0.4 mol H2SO4, u must have 0.4 / 3 = 2/15 mol of Fe(OH)3 the solution with the volume of 27.0 mL or 0.027 L, which contains 2/15 mol Fe(OH)3 will have the molar concentration of 2/15 / 0.027 = 4.938 (M)
From Youtube
Neutralization Reaction Practice Problem #4  Acid Base Reactions www.whitwellhigh.com :Neutralization Reaction Practice Problem #4  Acid Base Reactions  Chemistry  Whitwell High School  UTC  University of Tennessee at Chattanooga www.whitwellhigh.com Instructor/Professor: Johnny Cantrell
Neutralization Reaction Practice Problem #1  Acid Base Reactions www.whitwellhigh.com :Neutralization Reaction Practice Problem #1  Acid Base Reactions  Chemistry  Whitwell High School  UTC  University of Tennessee at Chattanooga www.whitwellhigh.com Instructor/Professor: Johnny Cantrell