Calculating pH during titration?
100.0 mL of 0.100 M acetic acid is placed in a flask and is titrated with 0.100 M sodium hydroxide. An appropriate indicator is used. Ka for acetic acid is 1.7 s 10^-5. Calculate the pH in the flask at the following points in the titration. a) When no NaOH has been added b) After 25.0 ml of NaOH is added c) After 50.0 ml of NaOH is added d) After 75.0 ml of NaOH is added e) After 100.0 ml of NaOH is added f) After 300.0 ml of NaOH is added I just want to make sure that my math is right... For A- I got a pH of 1 by doing the -log of ((.100 M x .100 L) - (.100 M x 0 L)) // .100 L. I used the same technique to get b) 1.22 c) 1.48 d) 1.85 Is that correct? And then I'm having a hard time with e because with what I'm doing I get -log(0) which of course is undefined.... Help would be appreciated!
Which of the following statements is true concerning a concentrated solution of lithium oxide in water?
Perform the chemical reaction using a balanced chemical equation and look at your product to determine what it is: Li2O + H2O ---> 2 LiOH Okay, so our product is 2 LiOH. Look at the molecular formula for LiOH. What is the anion? Hydroxide is the anion. Is hydroxide a strong or weak acid or base? Hydroxide is a strong base. Thus, A. strongly basic.
A solution contains 2.0 x 10 ^-4 M Ag+ and 1.5.....?
your first answer: AgI ppt's first =================== AgI K = [Ag] [I] 8.3e-17 = [2e-4][I] [I] = 4.2e-13 molar Your 2nd answer: to get AgI to ppt, [I] > 4.2e-13 Molar ------------------------ PbI2 K = [Pb] [I]^2 7.9e-9 = [1.5e-3] [I]^2 I2 = 5.26e-6 I = 2.3e-3 Molar lastr answer:Pb needs the [I] > 2.3e-3 Molar
Which statement is true concerning the solution process? HELP PLEASE!!!!?
the solution process can be thought of as a three step process where: (1) intermolecular interactions in the solute are broken, (2) solvent cavities are formed, and (3) solute particles enter solvent cavities. Which of the following statements is true concerning the solution process? a: Forming solvent cavities is endothermic, solute particles entering solvent cavities is endothermic b: Breaking the intermolecular interactions in the solute is endothermic; solute particles entering solvent cavities is exothermic c: Forming solvent cavities is exothermic, solute particles entering solvent cavities is endothermic d: Breaking the intermolecular interactions in the solute is endothermic, forming solvent cavities is exothermic
What happens when sodium hydroxide and copper sulfate are combined? Is the reaction dangerous?
NaOH is a strong base while CuSO4 is an acidic salt. When their solutions are mixed with each other, a pale blue precipitate of basic copper hydroxide & a solution of neutral salt sodium sulphate will be formed.C. E.2NaOH + CuSO4 --> Cu(OH)2 + Na2SO4.This is a double displacement as well as precipitation reaction. According to my knowledge, this reaction should not be so dangerous.
Calling All Chemistry Geniuses! Your Help Is Needed!?
1. 2A + 3B --> 5C 2. The answer is 1.0. moles = 1.2 x 10^24 atoms(1 molecule/2 atoms)(1 mole/6.022 x 10^23) = 1.0 3. 1.8 × 10^24 4. 23.95 g/mol(3 moles) = 72 g 5. - 236 g/mol 6. 22.4 L CO2/mole CO2(1 mole CO2/2 moles LiOH)(4 moles LiOH) = 44.4 moles -45 L 7. In 22.4 L at STP, all gases have 6.022 x 10^23 molecules. - The gases have the same number of molecules. 8. moles H2O = 16 moles C2H6(6 moles H2O/2 moles C2H6) -48 moles 9. L HCl = 11.2 L Cl2(1 mole Cl2/22.4 L Cl2)(2 mole HCl/1 mole Cl2)(22.4 L HCl/1 mole HCl) - 22.4 L 10. L CO2 = 269 L C2H6(1 mole C2H6/22.4 L C2H6)(4 moles CO2/2 moles C2H6)(22.4 L CO2/1 mole CO2) = 538 L CO2 -538 L 11. If a 102 g sheet of aluminum oxide formed completely in excess oxygen, how many grams of aluminum were oxidized? Trick question. -102 g Who's buried in Grant's tomb? 12. -A/B because you need three times as much B to have a stoichiometric reaction 13. -NaNO3 Actually it doesn't react with NaOH 14. 386.65 g/mol -386 g/mol 15.-A/3B 16. Theor. mol NH3 = 2 mol H2(2 mol NH3/3 mol H2) = 1.33 mol NH3 produced from 2 mol H2 % yield = (0.54 mol/1.33 mol) x 100% = 40.5% -40% 17. -The product participates in a side reaction. 18. LiOH 23.95 g/mol moles LiOH = 24 g LiOH(1 mol LiOH/23.95 g LiOH) = 1.00 mol LiOH mass Li3N = 1.00 mol LiOH(1 mol LiN3/3 mol LiOH)(34.83 g LiN3/1 mol LiN3) = 11.6 g -11.7 g There you go. Hope it helps.