Tony Spark 194 Questions 0 Answers 0 Best Answers 1k Points View Profile Tony SparkEnlightened Asked: April 25, 20202020-04-25T15:49:11+00:00 2020-04-25T15:49:11+00:00In: Chemistry questions and answers Write the balanced chemical equation for the reaction of KHP with NaOH. Suppose your laboratory i… Need Help! 2 Answers Voted Oldest Recent Anonymous Added an answer on April 25, 2020 at 3:49 pm Part 1) This is an acid and base reaction. KHP a weak acid reacts with a strong base sodium hydroxide to produce a weak base KNaP plus water.The reaction proceeds according to the equation: KHC8H4O4(aq) + NaOH(aq) –> KNaC8H4O4(aq) + H2O(l) Part 2) If the solution was contaminated by NaCl, There would be dissociation of NaCl into Na+ and Cl- which will increase the concentration of Na+ in the solution. Increase in concentraion of Na+ will hamper the calculation of Molarity of NaOH. Part 3) malonic acid is a diprotic acid so, for every 1 mole malonic acid, 2 moles NaOH are required to neutralize it. 0.1M NaOH x 0.02182L = 0.0021moles NaOH 0.0021moles NaOH x (1H2C3H2O4 / 2NaOH) / 0.01212L = 0.086M Part 4) First find the number of moles of sodium carbonate used. Moles = grams sodium carbonate over grams sodium carbonate per mole Moles = .432 g over l06 g/mole sodium carbonate = 4.08 x l0^-3 One mole of NaCO3 will neutralize 1 mole of H2SO4 . Molarity of H2SO4 = moles acid /LITERS solution Molarity = 4.08 x l0^-3 moles/ .0223 LITERS = 1.83 Molar H2SO4 Part 5) Molecular wt. of oxalic acid = 90.03 g/mole 0.252 g x (1 mole / 90.03 g) x ( 1 / 0.500L) = 0.0056 moles /L = 0.0056 M 0 Reply Share Share Share on Facebook Share on Twitter Share on LinkedIn Share on WhatsApp Anonymous Added an answer on April 25, 2020 at 3:49 pm 1.Potassium acid phthalate is a large molecule (KHC8H4O4) with a molar mass of 204.2 g/mol. Instead of writing the whole formula, we abbreviate it as KHP, where “P” stands for the phthalate ion, (C8H4O4)- – and not for phosphorus. KHP is an acidic substance, with the ionizing hydrogen being set forward in the formula for emphasis. Therefore, KHP is monoprotic and will react with NaOH in a simple 1 to 1 relationship according to the following equation: NaOH(aq) + KHC8H4O4 (aq) —> KNaC8H4O4 (aq) + H2O(l) 2.Assuming you mean potassium hydrogen phosphate, K2HPO4, the net ionic equation would be: (HPO4)- + (OH)- –> H2O + (PO4)3- with (HPO4)- and (OH)- reacting in a 1:1 ratio. Contamination with NaCl would mean that you actually weighed out less K2HPO4 than you thought. This would mean that the K2HPO4 solution was less concentrated than you thought. This would mean that a given volume of this solution (say 25cm3) would neutralise less NaOH than it should. Given that the concentration of NaOH would be calculated as: [NaOH] = (assumed moles of K2HPO4)/(volume of NaOH(aq)) and ‘volume of NaOH’ is too small, the value you would calculate would be too large 3.Malonic acid (HOOCCH2COOH) has 2 replaceable protons H2A + 2NaOH –> Na2A + 2H2O You require 2 moles of NaOH for every 1 mole of malonic acid 21.82 mL NaOH x 0.1000 mole/L = 2.182 millimoles of NaOH (0.002076 mole) Millimoles of malonic acid neutralized by 2.182 millimolesd of NaOH = 1.091 millimoles (2.182 / 2) 13.15 mL malonic acid x molarity = 1.091 millimoles molarity = 1.091 millimoles / 12.12 mL = 0.0900 M (moles/L or millimoles/mL) ANSWER: 0.0900 M 4.First find the number of moles of sodium carbonate used. Moles = grams sodium carbonate over grams sodium carbonate per mole Moles = .432 g over l06 g/mole sodium carbonate = 4.08 x l0^-3 Look at the formula for sodium carbonate and see that one mole of sodium carbonate will contain one mole of CO3 = ion, and since this ion has a charge of -2, it can neutralize two H+ ions. Therefore 4.08 x l0^-3 moles of CO3= will neutralize twice as many moles of H+ or 8.15 x l0^-3 moles of H+ And in one mole of H2SO4 there are two moles of H+ So to get moles of H2SO4 neutralized, we cut moles H+ in half to get 4.08 x l0^-3 moles sulfuric acid neutralized. Now Molarity sulfuric acid = moles acid over LITERS solution so Molarity = 4.08 x l0^-3 moles over .0223 LITERS (same as 22.3 ml) and Molarity of acid = l.83 Molar H2SO4 5.oxalic acid has molecular weight = 90.03 g/mole 0.252 g x (1 mole / 90.03 g) x ( 1 / 0.500L) = 0.0056 moles /L = 0.0056 M 0 Reply Share Share Share on Facebook Share on Twitter Share on LinkedIn Share on WhatsApp Leave an answerLeave an answerCancel reply Featured image Select file Browse Answer Anonymously Save my name, email, and website in this browser for the next time I comment.