average molarity of NaOH (from part A), M 0.5507 m NaDH HC,H,0,(aq) + NaOH(aq) -→ H,O(1) + NaC,H,O,(aq) Trial 1 Trial 2 volume of vinegar, mL 10.00 mL 10.00 mL volume of vinegar, L initial buret reading, mL 0.00 mL 0.00mL final buret reading, mL 38.50 mL るい50mL volume of NaOH, mL 38.50mh 36.50mL volume of NaOH, L 0.0383 mL 0.03し5 レ moles of NaOH, mol moles of HC,H,O,(aq), mol (HC,H,0,(aq)], M average [HC,H,O,(aq)], M
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I need help completing the missing pieces on this table for trial 1 & 2
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- A titration of H2SO4 with NaOH. They finds that it takes 19.87 mLof 0.2091 M NaOH to reach the endpoing using 15.00 mL of H2SO4. Use this informationto calculate the molarity of the H2SO4 H2SO4(aq)+2NaOH(aq)-->Na2SO4(aq)+2H2O(l)A 50.0-mL solution of 0.100 M propanoic acid (HC3H5O2; Ka= 1.34 x 10–5) is titrated with 0.200 M NaOH solution. The net ionic equation for the reaction is as follows: HC3H5O2(aq) + OH–(aq) --> C3H5O2–(aq)+ H2O (a) What is the initial pH of the acid solution? (b) Calculate the pH of the resulting solution after each of the following amount of 0.200 M NaOH has been added, and the limiting reactant is completely reacted. (i) 10.0 mL of 0.200 M NaOH; (ii) 15.0 mL of 0.100 M NaOH; (iii) 20.0 mL of 0.100 M NaOH; (iv) 25.0 mL of 0.100 M NaOH; (v) 30.0 mL of 0.200 M NaOHA 50 mL portion of an HCL solution requires 29.71 mL of 0.01963 M Ba(OH)2 to reach an endpoint with bromocresol green indicator. Calculate the molarity of the HCl ?
- . A 4.59 mL sample of HCl, specific gravity 1.3, required 50.5 mL of 0.9544N NaOH in a titration. Calculate the % w/w HCl.Part C. Determination of Unknown Balanced chemical equation: Hcl+ Na OH -o Nacl t Hg0 Unknown number NA Trial 1 Trial 2 Trial 3 Volume of unknown HCl titrated 10.00mL NA NA Initial burette reading 0.90mL Final burette reading 30.90mL Volume NaOH added 20 00ML 50.00MLXO.144 m=43.2m moles NaOH added 4.335m (Use average molarity from part B and multiply it by the volume you added) 0.4335 M moles HCl 4.335m 04335m 10.00 mL molarity of HCl Actual Molarity (from instructor) 0.3150M percentage errorWhat is the Qsp of a solution mixture Al(OH)3 formed from 10 mL 1.00M C6H5NH2 and 15 mL 0.10M Al(NO3)3?
- A 0.098 M K2Cr2O7 titrant was used for titration of 30 mL of aqueous Ethanol (C2H6O) solution and consumed 41.40 mL. (a) Calculate the mass of Ethanol and (b) determine degree proof of ethanol (density of pure ethanol= 0.789 g/mL).How many milliliters of 8.50x10-2 M NaOH are required to titrate each of the following solutions to the equivalence point? Show Transcribed Text V = 45.0 mL of 9.00x10-² M HNO3 Submit ▾ Part B V = —| ΑΣΦ 35.0 mL of 8.50x10-2 M CH3COOH Show Transcribed Text V = Request Answer IVE ΑΣΦ P Pearson Submit C IVE ΑΣΦ Ĉ Request Answer 55.0 mL of a solution that contains 1.80 g of HCl per liter www ? SW ? ? mL mL mL3.) Generate a titration curve for 25.0 ml of 0.7233 M H;PO4, with 1.256 M NaOH as the titrant, given the following total volumes of titrant dispensed. H3PO4 = H2PO: Ko1 = 7.11 x 10 3 Koz = 6.32 x 10-3 Ke3 = 4.50 x 10-13 H2PO, = HPO÷² HPO:? = PO43 V NAOH (mL): 0.00, 2.0, 12.0, 20.0, 30.0, 38.0, 45.0, 57.0, 66.0, 70.0, 85.0
- A 0.492-g sample of KH2PO4 is titrated with 0.112 M NaOH, requiring 25.6 mL: H2PO4- + OH- → HPO42- + H2O What is the percent purity of the KH2PO4 (FW = 136.09)?Calculate the volume in milliliters of 0.368 M KOH necessary to titrate 0.0209 moles of acetic acid (HOAc, HC2H3O2) to a phenolphthalein end-point. Report the answer with a precision of two decimal places.For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%. Is Kc > 1, < 1, or ≈ 1 for a titration reaction? Kc is the product of all reaction products (choose from: Multiplied by, added to, divided by, subtracted from) the product of the reactants, with all concentrations of reactants and products raised to their respective stoichiometric powers. As the titration proceeds, the amount of reactants (choose from: Increases, decreases, remains the same) and the amount of products (choose from: Increases, decreases, remains the same). Because the value of the numerator is (Choose from much greater than, much less than, approximately equal to) the value of the denominator, the value of Kc will be (Choose from much greater than, much less than, approximately equal to)1 for an effective titration reaction.