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Calculate the percent ionizationof a 0.15 M benzoic acid solution in pure water Nd also in a solution containing 0.10M sodium benzoate. Why does the percent ionization differ significantly in the two solutions?
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Benzoic acid reacts with water to form the benzoate ion by the following reaction C6H5COOH(aq) + H2O(l) ⇀↽ C6H5COO−(aq) + H3O +(aq) The equilibrium constant for this reaction is 6.4 x 10−5 . In a 0.1 M solution of benzoic
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Cyanic Acid is a weak acid HOCN + H2O H3O+ +OCN- ka= 3.5*10^-4 a) if 2.5ml of 0.01 M cyanic acid solution is added to 25.0 ml of a formis acid buffer with ph = 3.70, what is the ratio of [OCN-]/[HOCN] in the resulting solution? b)
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Given acidity constant,Ka of benzoic acid is 6.28 X 10^-5. pH of 0.15 molar solution of this acid is ? Here is my method: Benzoic acid is a weak acid,hence it dissociates very little. So: C6H5COOH---> C6H5COO- + H+ [H+] and
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1)Why benzoic acid is soluble in NaOH? 2)Why HCl is insoluble in titration of benzoic acid and NaOH? Benzoic is soluble in a solution of NaOH because the base forms the sodium salt with the acid to form sodium benzoate. The sodium
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A 0.150 M solution of nitrous acid (HNO2) is made. Ka = 4.5 x 10-4 1. Show the equilibrium which occurs when this acid is dissolved in water. 2. What is the pH of the solution? Show all work clearly. 3. 100.0 mL of the solution is
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The enthalpy of combustion of benzoic acid(C6H5COOH) which is often used to calibratecalorimeters, is −3227 kJ/mol. When 1.316 gof benzoic acid was burned in a calorimeter, the temperature increased by 2.226◦C. What is the
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1. Calculate the pH of a buffer solution that contains 0.32 M benzoic acid (C6H5CO2H) and 0.17 M sodium benzoate (C6H5COONa). [Ka = 6.5 × 10-5 for benzoic acid] Round your answer to two places past the decimal. 2. A solution is