PRE-LAB QUESTIONS: NAME 1. Write the complete balanced molecular equation for th

Question

PRE-LAB QUESTIONS: NAME 1. Write the complete balanced molecular equation for the rea between aaneous susuric acid and aqueous potassium hydroxide write also the ionic net equations for this reaction use parentheses to show the physical state of each reactant and product e Net Ionic 2 A 2050g sample of potassium hydrogen suteate was dissolved in distiled and then titrated to the endpoint with 20-25 mL of a potassium hydroxide solution. What is the concentration of the potassium hydroxide solution? The balanced chemical equation is Kasodaq) H 3. A sample weighing 1.731 g a mixture of the triprotic citric acid, H.csHso, and sodium sulfate, contains Nazsoe The sample mixture was dissolved in water and then titrated wth a 0.3550 M sodium hydroxide solution. It required 28.32 mL of the base to completely neutralize the citric acid. What is the percent of sodium sulfate in the mixture? (Note that Naso is chemically inert toward sodium hydroxide, i.e. it will not react with any of the sodium hydroxide used in the titration.) 4. A o 2046 g sample of a diprotic acid of the general formula HA required 18.55 mL of a 0.1040 M NaOH solution to neutralize it in an acid-base titration. What is the molar mass of the diprotic acid?

Explanation / Answer

Answer

1) Molecular Equation: H2SO4(aq) + 2KOH(aq) ----> K2SO4(aq) + 2H2O(l)

Ionic Equation: H+ + SO42- + K+ + OH- -----> K+ + SO42- + H2O

Net Equation: H+ + OH-  -----> H2O

2) The volume of distilled water in which KHSO4 is dissolved is not given. Assuming it is 50mL.

[KHSO4] = (0.2050/136.169)*(1000/50) = 0.0301 M

[KOH] = (0.0301*50)/20.25 = 0.0743 M

3) The volume of distilled water in which the mixture is dissolved is not given. Assuming it is 50mL.

[Citric acid] = 1/3*(28.32*0.3550/50) = 0.0670 M or 12.877 g/L or 0.6438 g/50mL

Wt. fraction of Na2SO4 = 1- (0.6438/1.731) = 62.806%

4) The volume of distilled water in which the acid is dissolved, is not given. Assuming it is 50mL.

[acid] = 1/2(18.55*0.1040/50) = 0.01929 M or 9.646 x 10-4 moles/50mL

Therefore the Molar mass of Acid = 0.2046/9.646 x 10-4 = 212.109 g/mol

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