Calculate the pH during the titration of 20.00 mL of 0.1000 M ammonia, NH3(aq), with 0.1000 M HBr(aq) after 18.97 mL of the acid have been added. Kb of ammonia = 1.8 x 10-5. (value + 0.02)
Q: 0.400 M
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- What mass of Ba(OH)2 is present in a sample if it is titrated to its equivalence point with 44.20 mL of 0.1000 N H2SO4? Note: Please present complete solution. Express your final answers up to two (2) decimal places.Molarity of (NH3) solution (M) from bottle- 5.0 Initial reading of buret (NH3) (mL)- .27 Final reading of buret (NH3) (mL)- 8.25 Volume of Cd(NO3)2 solution (mL)- 10.00 Volume of Na2C2O4 solution (mL)- 10,00 please find Total volume of solution after titration (mL) Total moles of C2O42- (mol) Molarity of C2O42- (M) Total moles of Cd2+ (mol) Moles of [Cd(NH3)4]2+ (mol) Molarity of [Cd(NH3)4]2+ (M) Moles of NH3 added by titration (mol) Moles of NH3 that did not react with Cd2+ (mol) Molarity of NH3 that did not react with Cd2+ (M) Kf for [Cd(NH3)4]2+Question 6 Calculate the pH during the titration of 20.00 mL of 0.1000 M butanoic acid (Ka = 1.54 x 10-5) with 0.1000 M sodium hydroxide solution after the addition of 20.00 mL of sodium hydroxide. correct sig fig and in decimals.
- Molarity of titrant (NaOH): 0.4550 M HC2H3O2 (aq) + NaOH (aq) → NaC2H3O2 (aq) + H2O (l) Trial # First Second Third Fourth Initial buret reading 0.15 mL 2.43 mL 1.32 mL 0.58 mL Final buret reading 18.62 mL 20.87 mL 20.03 mL 19.14 mL Volume of titrant used 18.47 mL 18.44 mL 18.71 mL 18.56 mL 4) Calculate the molarity of the acetic acid in the vinegar solution (Show your work). use FW for moles-->grams acetic acid. Molarity acetic acid = _____________ M 5) Calculate the weight % of acetic acid in the vinegar. How does this compare with the % listed on the label (5.00%)? (For this calculation assume that density of vinegar is 1.03 g/mL and of course, show your work). Weight % = ___________ 6) If you didn’t get the same weight % of acetic acid as listed on the vinegar label (5.00 %), what are two things (be specific) that could’ve happened during the experiment that could explain the variation from the expected weight %? To do…A sodium hydroxide (NaOH) solution was standardized with KHP primary standard. The concentration was found to be 0.1022 M. What is the concentration of a 25.0-ml hydrochloric acid (HCl) solution if it required 32.6 ml of NaOH to reach the phenolphthalein end point?A 21.3 mL solution of HBr (a strong acid) of unknown concentration was titrated using 0.2735 mol L-1 NaOH using phenolphthalein as a pH indicator. At the moment when the solution turned a consistent light pink colour, the burette volume read 29.78 mL. The initial burette reading before the experiment was 7.15 mL. What was the concentration of HBr in the original HBr solution in mol L-1 ?
- If a 1.065 g sample of magnesium oxide of 84.736% were treated with 50 mL of 1.017 N sulfuric acid, what volume of 1.103 N sodium hydroxide would be required in the back titration? Sub question #1: What is the amount (in mg) of the analyte that is equivalent to one milliliter of the titrant at its equivalence point?Calculate the pH when 0.040 mL, 0.5 M sodium hydroxide titrant is added to 30 mL, 0.5 M acetic acid. Show pertinent solutions. Ka = 1.76 x 10-5A student titrated a 5.00-mL saturated borax sample at 15.7°C. The titration reaction is as follows: B4O5(OH)42–(aq) + 2 HCl(aq) + 3 H2O(l) → 4 B(OH)3(aq) + 2 Cl–(aq) What is the KSP of borax (Na2B4O5(OH)4) if the titration required 8.26 mL of 0.2004 M HCl?
- Question: The titration of a 25.0 mL sample of an aqueous solution containing HNO3 requires 18.6 mL of 0.150 M Ba(OH)2 to reach the equivalence point. The concentration of HNO3 in the original solution was ______ M. What I did: c=n/V n=0.150M * 18.6 mL * (10^-3 L)/1 mL = 0.00279 mol c= 0.00279 mol/25.0 mL * 1 mL/(10^-3 L) = 0.1116 M --> 0.112M It says it's wrong but I think the wording just confused me. I think I have the right idea though but I just don't know where to go with this. How can I fix my math so that I have the correct answer?QUESTION 12 Determine the volume in mL of 0.181 M HCl(aq) needed to reach the equivalence (stoichiometric) point in the titration of 37.23 mL of 0.175 M NH2NH2(aq)(aq). The Kb of hydrazine is 1.7 x 10-6. (value ± 1%)Na2CO3 served as the primary standard in a titration experiment. Find the molarity of the titrant given the following data in 3 decimal places. Show solutions Primary Standard Used: Na2CO3Formula Mass of 1º standard: 105.99 g/mol% purity of 1º standard: 95% Trial 1 2 3 1º Standard weight, g 0.1005 0.1001 0.0997 Net volume of HCl, mL 9.30 9.00 8.90 Molarity of HCl X1 X2 X3